Linear differential equations Questions in English

(C) Given the equation $\int f(x) \, dx = x e^{-\log |x|} + f(x)$.
We know that $e^{-\log |x|} = e^{\log |x|^{-1}} = |x|^{-1} = \frac{1}{|x|}$.
Substituting this into the equation,we get $\int f(x) \, dx = x \cdot \frac{1}{|x|} + f(x)$.
Assuming $x > 0$,$|x| = x$,so $\int f(x) \, dx = 1 + f(x)$.
Differentiating both sides with respect to $x$,we get $f(x) = 0 + f'(x)$,which implies $f'(x) = f(x)$.
This is a first-order linear differential equation $\frac{df}{dx} = f(x)$,which can be written as $\frac{df}{f} = dx$.
Integrating both sides,we get $\log |f(x)| = x + C$,which implies $f(x) = e^{x+C} = c e^x$.

(A) Given the differential equation: $\frac{dy}{dx} = \sec x(\sec x + \tan x)$
Expanding the right side,we get: $\frac{dy}{dx} = \sec^2 x + \sec x \tan x$
To find $y$,we integrate both sides with respect to $x$:
$y = \int (\sec^2 x + \sec x \tan x) dx$
Using the standard integrals $\int \sec^2 x dx = \tan x$ and $\int \sec x \tan x dx = \sec x$,we get:
$y = \tan x + \sec x + c$

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -1$ and $Q(x) = 1$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.
Multiplying both sides by the $IF$,we get $e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x}$,which simplifies to $\frac{d}{dx}(y e^{-x}) = e^{-x}$.
Integrating both sides with respect to $x$,we get $y e^{-x} = \int e^{-x} dx = -e^{-x} + C$.
Thus,$y = -1 + C e^x$.
Using the initial condition $y(0) = -1$,we substitute $x = 0$ and $y = -1$:
$-1 = -1 + C e^0 \Rightarrow -1 = -1 + C \Rightarrow C = 0$.
Substituting $C = 0$ back into the general solution,we get $y(x) = -1$.

(C) differential equation is linear if the dependent variable $y$ and its derivatives appear only in the first degree and are not multiplied together.
In option $A$,the term $(\frac{d^2y}{dx^2})^2$ has a degree of $2$,so it is non-linear.
In option $B$,the term $\sqrt{1 + (\frac{dy}{dx})^2}$ makes the equation non-linear because the derivative is inside a square root.
In option $C$,$\frac{dy}{dx} + \frac{y}{x} = \log x$ is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \log x$. This is a linear differential equation.
In option $D$,the term $y\frac{dy}{dx}$ involves the product of the dependent variable and its derivative,making it non-linear.
Therefore,the correct option is $C$.

(A) This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = x^m \cos x$.
First,we calculate the integrating factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \cdot \sec x = \int (x^m \cos x) \cdot \sec x dx + c$.
Since $\cos x \cdot \sec x = 1$,the equation simplifies to:
$y \sec x = \int x^m dx + c$.
Integrating $x^m$ with respect to $x$:
$y \sec x = \frac{x^{m + 1}}{m + 1} + c$.
Multiplying both sides by $(m + 1) \cos x$:
$(m + 1) y = x^{m + 1} \cos x + c(m + 1) \cos x$.

(B) Given the differential equation: $(1 + y^2)dx - (\tan^{-1} y - x)dy = 0$
Rearranging the terms to form a linear differential equation in $x$:
$(1 + y^2)dx = (\tan^{-1} y - x)dy$
$\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1 + y^2}$
$\frac{dx}{dy} = \frac{\tan^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$
Adding $\frac{x}{1 + y^2}$ to both sides:
$\frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{\tan^{-1} y}{1 + y^2}$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{1 + y^2}$ and $Q = \frac{\tan^{-1} y}{1 + y^2}$.
The integrating factor $(I.F.)$ is given by:
$I.F. = e^{\int P dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y}$.

(A) The given differential equation is $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$.
Dividing by $(1 + {x^2})$,we get $\frac{{dy}}{{dx}} + \frac{{2x}}{{1 + {x^2}}}y = \frac{{4{x^2}}}{{1 + {x^2}}}$.
This is a linear differential equation of the form $\frac{{dy}}{{dx}} + Py = Q$,where $P = \frac{{2x}}{{1 + {x^2}}}$ and $Q = \frac{{4{x^2}}}{{1 + {x^2}}}$.
The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values,$y(1 + {x^2}) = \int \frac{{4{x^2}}}{{1 + {x^2}}} (1 + {x^2}) dx + c$.
$y(1 + {x^2}) = \int 4{x^2} dx + c = \frac{{4{x^3}}}{3} + c$.
Since the curve passes through the origin $(0,0)$,we substitute $x=0$ and $y=0$: $0(1+0) = 0 + c$,which gives $c = 0$.
Thus,the equation of the curve is $y(1 + {x^2}) = \frac{{4{x^3}}}{3}$,or $3y(1 + {x^2}) = 4{x^3}$.

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = x^2$.
First,we find the integrating factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is given by $y \times (I.F.) = \int (Q \times I.F.) dx + c$.
Substituting the values,we get $y \times x = \int (x^2 \times x) dx + c$.
$xy = \int x^3 dx + c$.
$xy = \frac{x^4}{4} + c$.
Multiplying both sides by $4$,we get $4xy = x^4 + 4c$.
Since $4c$ is an arbitrary constant,we can write it as $c$.
Thus,the solution is $4xy = x^4 + c$.

(A) The given differential equation is $x\frac{dy}{dx} + y = x^2 + 3x + 2$.
Dividing both sides by $x$,we get $\frac{dy}{dx} + \frac{1}{x}y = x + 3 + \frac{2}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = x + 3 + \frac{2}{x}$.
The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The solution is given by $y(I.F.) = \int Q(I.F.) dx + c$.
Substituting the values,$y(x) = \int (x + 3 + \frac{2}{x})x dx + c$.
$xy = \int (x^2 + 3x + 2) dx + c$.
$xy = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + c$.

(D) The given differential equation is $\frac{dy}{dx} + \frac{3x^2}{1 + x^3}y = \frac{\sin^2 x}{1 + x^3}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1 + x^3}$ and $Q = \frac{\sin^2 x}{1 + x^3}$.
First,we find the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \frac{3x^2}{1 + x^3} dx} = e^{\ln(1 + x^3)} = 1 + x^3$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y(1 + x^3) = \int \frac{\sin^2 x}{1 + x^3} \cdot (1 + x^3) dx + c$
$y(1 + x^3) = \int \sin^2 x dx + c$
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:
$y(1 + x^3) = \int \frac{1 - \cos 2x}{2} dx + c$
$y(1 + x^3) = \frac{1}{2} \int (1 - \cos 2x) dx + c$
$y(1 + x^3) = \frac{1}{2} (x - \frac{\sin 2x}{2}) + c$
$y(1 + x^3) = \frac{x}{2} - \frac{\sin 2x}{4} + c$.

(B) differential equation is linear if the dependent variable $y$ and its derivative $\frac{dy}{dx}$ occur only in the first degree and are not multiplied together.
Option $(b)$ is $x^2\frac{dy}{dx} + y = e^x$.
Dividing by $x^2$,we get $\frac{dy}{dx} + \frac{1}{x^2}y = \frac{e^x}{x^2}$.
This is in the standard form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x^2}$ and $Q = \frac{e^x}{x^2}$.
Therefore,it is a linear differential equation.

(B) Given the linear differential equation: $x\frac{dy}{dx} + 3y = x$.
Divide the entire equation by $x$ to get it into the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{3}{x}y = 1$.
Here,$P = \frac{3}{x}$ and $Q = 1$.
Calculate the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln|x|} = e^{\ln|x^3|} = x^3$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \cdot x^3 = \int 1 \cdot x^3 dx + c$.
$y x^3 = \frac{x^4}{4} + c$.

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = \cos x$.
First,we find the integrating factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values,we get $y e^x = \int \cos x \cdot e^x dx + c$.
To solve $\int e^x \cos x dx$,we use the integration by parts formula $\int u v dx = u \int v dx - \int (u' \int v dx) dx$:
Let $I = \int e^x \cos x dx$.
$I = e^x \cos x - \int e^x (-\sin x) dx = e^x \cos x + \int e^x \sin x dx$.
Using integration by parts again for $\int e^x \sin x dx$:
$I = e^x \cos x + (e^x \sin x - \int e^x \cos x dx) = e^x \cos x + e^x \sin x - I$.
$2I = e^x(\cos x + \sin x) \implies I = \frac{1}{2} e^x(\cos x + \sin x)$.
Thus,$y e^x = \frac{1}{2} e^x(\cos x + \sin x) + c$.
Dividing by $e^x$,we get $y = \frac{1}{2}(\cos x + \sin x) + ce^{-x}$.

(D) The given differential equation is $\frac{dy}{dx} + y \cot x = 2 \cos x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 2 \cos x$.
First,we find the integrating factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values,we get $y \sin x = \int (2 \cos x \sin x) dx + c$.
Using the identity $2 \sin x \cos x = \sin 2x$,we have $y \sin x = \int \sin 2x dx + c$.
Integrating $\sin 2x$,we get $y \sin x = -\frac{\cos 2x}{2} + c$.
Multiplying by $2$,we obtain $2y \sin x = -\cos 2x + 2c$.
Rearranging the terms,we get $2y \sin x + \cos 2x = C$ (where $C = 2c$).
Thus,the correct option is $D$.

(B) Given the equation $(x + 2y^3)\frac{dy}{dx} - y = 0$.
Rearranging the terms,we get $(x + 2y^3)\frac{dy}{dx} = y$.
This can be written as $\frac{dx}{dy} = \frac{x + 2y^3}{y}$.
Simplifying,we get $\frac{dx}{dy} = \frac{x}{y} + 2y^2$,or $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = 2y^2$.
The integrating factor $(I.F.)$ is given by $e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.
The general solution is $x \cdot (I.F.) = \int Q \cdot (I.F.) dy + A$.
Substituting the values,$x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + A$.
$x \cdot \frac{1}{y} = \int 2y dy + A$.
$x \cdot \frac{1}{y} = y^2 + A$.
Multiplying by $y$,we get $x = y^3 + Ay$,which can be rewritten as $y^3 - x = -Ay$. Since $A$ is an arbitrary constant,we can write the solution as $y^3 - x = Ay$.

(B) The given differential equation is $\frac{dy}{dx} = y \tan x - y^2 \sec x$.
This is a Bernoulli's equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$,where $n = 2$.
Dividing both sides by $y^2$,we get:
$y^{-2} \frac{dy}{dx} - y^{-1} \tan x = - \sec x$ ..... $(i)$
Let $v = y^{-1} = \frac{1}{y}$. Then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,which implies $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.
Substituting these into equation $(i)$,we get:
$-\frac{dv}{dx} - v \tan x = - \sec x$
Multiplying by $-1$,we obtain the linear differential equation:
$\frac{dv}{dx} + v \tan x = \sec x$
Comparing this with the standard linear form $\frac{dv}{dx} + P(x)v = Q(x)$,we have $P(x) = \tan x$.
The integrating factor $(I.F.)$ is given by:
$I.F. = e^{\int P(x) dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
Thus,the correct option is $B$.

(C) Given differential equation is $\cos x \frac{dy}{dx} + y \sin x = 1$.
Dividing both sides by $\cos x$,we get:
$\frac{dy}{dx} + y \tan x = \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sec x$.
The integrating factor $(I.F.)$ is given by $e^{\int P \, dx}$.
$I.F. = e^{\int \tan x \, dx} = e^{\ln |\sec x|} = \sec x$.
Thus,the correct option is $C$.

(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = a$ and $Q = e^{mx}$.
The Integrating Factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int a dx} = e^{ax}$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
Substituting the values,we get $y \cdot e^{ax} = \int e^{mx} \cdot e^{ax} dx + C$.
$y \cdot e^{ax} = \int e^{(a+m)x} dx + C$.
$y \cdot e^{ax} = \frac{e^{(a+m)x}}{a+m} + C$.
Dividing both sides by $e^{ax}$,we get $y = \frac{e^{mx}}{a+m} + C e^{-ax}$.
Multiplying by $(a+m)$,we get $(a+m)y = e^{mx} + C(a+m)e^{-ax}$.
Thus,the correct option is $(d)$.

(C) The given differential equation is $\frac{dy}{dx} + y \tan x = \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sec x$.
The integrating factor $(I.F.)$ is given by the formula $I.F. = e^{\int P \, dx}$.
Substituting $P = \tan x$,we get $I.F. = e^{\int \tan x \, dx}$.
Since $\int \tan x \, dx = \ln|\sec x|$,we have $I.F. = e^{\ln(\sec x)}$.
Using the property $e^{\ln(f(x))} = f(x)$,we get $I.F. = \sec x$.
Since $\sec x = \frac{1}{\cos x}$,the correct option is $C$.

(C) Given differential equation is $x \log x \frac{dy}{dx} + y = 2 \log x$.
Dividing both sides by $x \log x$,we get:
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$.
Integrating factor $(I.F.) = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx}$.
Let $\log x = t$,then $\frac{1}{x} dx = dt$.
So,$I.F. = e^{\int \frac{1}{t} dt} = e^{\log t} = t = \log x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
$y \log x = \int \frac{2}{x} \cdot \log x dx$.
Let $\log x = u$,then $\frac{1}{x} dx = du$.
$y \log x = \int 2u du = u^2 + c$.
Substituting $u = \log x$,we get $y \log x = (\log x)^2 + c$.

(A) The given differential equation is $\frac{dy}{dx} + (2 \cot x)y = 3x^2 \csc^2 x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 \cot x$ and $Q = 3x^2 \csc^2 x$.
First,we find the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int 2 \cot x dx} = e^{2 \ln|\sin x|} = e^{\ln(\sin^2 x)} = \sin^2 x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \cdot \sin^2 x = \int (3x^2 \csc^2 x) \cdot \sin^2 x dx$.
Since $\csc^2 x = \frac{1}{\sin^2 x}$,the expression simplifies to:
$y \sin^2 x = \int 3x^2 dx$.
Integrating $3x^2$ with respect to $x$ gives $x^3$.
Thus,$y \sin^2 x = x^3 + c$.

(C) Given the differential equation: $x \frac{dy}{dx} = y + x^2$.
Dividing both sides by $x$ (assuming $x \neq 0$): $\frac{dy}{dx} - \frac{1}{x} y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = x$.
The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\log_e x} = e^{\log_e (x^{-1})} = \frac{1}{x}$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + a$.
Substituting the values: $y \cdot \frac{1}{x} = \int x \cdot \frac{1}{x} dx + a$.
$\frac{y}{x} = \int 1 dx + a$.
$\frac{y}{x} = x + a$.
Therefore,$y = x^2 + ax$.

(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = x^3 - 3$.
The integrating factor $(I.F.)$ is given by the formula $I.F. = e^{\int P dx}$.
Substituting the value of $P$,we get $I.F. = e^{\int \frac{1}{x} dx}$.
Since $\int \frac{1}{x} dx = \log_e x$,we have $I.F. = e^{\log_e x}$.
Using the property $e^{\log_e x} = x$,the integrating factor is $x$.

(B) Given the differential equation: $\cos x \frac{dy}{dx} + y \sin x = 1$
Dividing both sides by $\cos x$,we get: $\frac{dy}{dx} + y \tan x = \sec x$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sec x$.
The Integrating Factor $(I.F.)$ is given by: $I.F. = e^{\int P dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The general solution is given by: $y \times (I.F.) = \int (Q \times I.F.) dx + c$
Substituting the values: $y \sec x = \int (\sec x \times \sec x) dx + c$
$y \sec x = \int \sec^2 x dx + c$
$y \sec x = \tan x + c$.

(B) The given equation is $\frac{dy}{dx} + 2y \tan x = \sin x$,which is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 \tan x$ and $Q = \sin x$.
First,we find the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = e^{\ln |\sec^2 x|} = \sec^2 x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \sec^2 x = \int \sin x \cdot \sec^2 x dx + c$.
Since $\sin x \cdot \sec^2 x = \sin x \cdot \frac{1}{\cos^2 x} = \tan x \sec x$,we have:
$y \sec^2 x = \int \tan x \sec x dx + c$.
Integrating $\int \tan x \sec x dx$,we get $\sec x$.
Thus,the solution is $y \sec^2 x = \sec x + c$.

(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \sec^2 x$ and $Q = \tan x \sec^2 x$.
First,we find the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P \, dx} = e^{\int \sec^2 x \, dx} = e^{\tan x}$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c$.
Substituting the values:
$y e^{\tan x} = \int (\tan x \sec^2 x) e^{\tan x} \, dx + c$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
The integral becomes $\int u e^u \, du$.
Using integration by parts: $\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u = e^u(u - 1)$.
Substituting back $u = \tan x$:
$y e^{\tan x} = e^{\tan x}(\tan x - 1) + c$.
Dividing both sides by $e^{\tan x}$:
$y = \tan x - 1 + c e^{-\tan x}$.

(C) The given differential equation is $(1 - x^2)\frac{dy}{dx} - xy = 1$.
Dividing both sides by $(1 - x^2)$,we get $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{x}{1 - x^2}$ and $Q = \frac{1}{1 - x^2}$.
The integrating factor $(I.F.)$ is given by $e^{\int P dx}$.
$I.F. = e^{\int -\frac{x}{1 - x^2} dx}$.
Let $u = 1 - x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
$I.F. = e^{\int \frac{1}{2u} du} = e^{\frac{1}{2}\log|u|} = e^{\log|u|^{1/2}} = \sqrt{1 - x^2}$.

(A) The given differential equation is $(x^2 + 1)\frac{dy}{dx} + 2xy = x^2 - 1$.
Dividing both sides by $(x^2 + 1)$,we get:
$\frac{dy}{dx} + \frac{2x}{x^2 + 1}y = \frac{x^2 - 1}{x^2 + 1}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{x^2 + 1}$ and $Q = \frac{x^2 - 1}{x^2 + 1}$.
The integrating factor $(I.F.)$ is given by $e^{\int P dx}$.
$I.F. = e^{\int \frac{2x}{x^2 + 1} dx}$.
Let $u = x^2 + 1$,then $du = 2x dx$.
$I.F. = e^{\int \frac{1}{u} du} = e^{\ln(u)} = u = x^2 + 1$.
Thus,the integrating factor is $x^2 + 1$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{3}$ and $Q = 1$.
First,we find the integrating factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \cdot e^{x/3} = \int 1 \cdot e^{x/3} dx + c$.
Integrating the right side:
$y \cdot e^{x/3} = 3e^{x/3} + c$.
Dividing both sides by $e^{x/3}$:
$y = 3 + c e^{-x/3}$.

(C) The given equation is $\frac{dy}{dx} + p(x)y = 0$.
This is a first-order linear differential equation.
We can solve this by separating the variables:
$\frac{dy}{y} = -p(x) dx$.
Integrating both sides,we get:
$\int \frac{dy}{y} = -\int p(x) dx$.
$\ln|y| = -\int p(x) dx + C_1$.
Taking the exponential of both sides:
$|y| = e^{-\int p(x) dx + C_1} = e^{C_1} e^{-\int p(x) dx}$.
Let $e^{C_1} = c$,then the solution is $y = c e^{-\int p(x) dx}$.

(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^{-x}$.
First,find the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is given by $y \times (I.F.) = \int (Q \times I.F.) dx + c$.
Substituting the values:
$y \times e^x = \int (e^{-x} \times e^x) dx + c$
$y e^x = \int 1 dx + c$
$y e^x = x + c$.
Using the initial condition $y(0) = 0$:
$0 \times e^0 = 0 + c \implies c = 0$.
Thus,the particular solution is $y e^x = x$,which simplifies to $y = x e^{-x}$.

(C) For a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x)$ and $Q(x)$ are functions of $x$ only,the integrating factor $(IF)$ is defined as:
$IF = e^{\int P(x) \, dx}$.
Thus,the correct option is $C$.

(B) The given equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$,which is known as Bernoulli's equation.
To linearize this equation,we divide both sides by $y^n$:
$y^{-n} \frac{dy}{dx} + P(x)y^{1-n} = Q(x)$
Now,we substitute $v = y^{1-n} = \frac{1}{y^{n-1}}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dv}{dx} = (1-n)y^{-n} \frac{dy}{dx}$
$\frac{1}{1-n} \frac{dv}{dx} = y^{-n} \frac{dy}{dx}$
Substituting these into the equation,we get:
$\frac{1}{1-n} \frac{dv}{dx} + P(x)v = Q(x)$
Multiplying by $(1-n)$,we obtain:
$\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$
This is a linear differential equation in $v$. Thus,the required substitution is $v = \frac{1}{y^{n-1}}$.

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = 1$.
First,we find the Integrating Factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \cdot e^x = \int 1 \cdot e^x dx + c$.
Integrating the right side:
$y \cdot e^x = e^x + c$.
Dividing both sides by $e^x$:
$y = \frac{e^x}{e^x} + \frac{c}{e^x} = 1 + c{e^{-x}}$.
Thus,the correct option is $A$.

(A) Given the differential equation: $dy = \cos x(2 - y \csc x)dx$
Dividing by $dx$,we get: $\frac{dy}{dx} = 2 \cos x - y \cot x$
Rearranging the terms: $\frac{dy}{dx} + y \cot x = 2 \cos x$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 2 \cos x$.
The integrating factor $(I.F.)$ is given by $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values: $y \sin x = \int (2 \cos x \cdot \sin x) dx + c = \int \sin(2x) dx + c$.
$y \sin x = -\frac{\cos(2x)}{2} + c$.
Alternatively,using $2 \cos x \sin x = \sin(2x)$ or integrating $2 \sin x \cos x$ directly gives $y \sin x = \sin^2 x + c$.
Given $y = 2$ when $x = \frac{\pi}{2}$:
$2 \sin(\frac{\pi}{2}) = \sin^2(\frac{\pi}{2}) + c \implies 2(1) = (1)^2 + c \implies c = 1$.
Thus,$y \sin x = \sin^2 x + 1$.
Dividing by $\sin x$,we get $y = \sin x + \csc x$.

(B) The given differential equation is $x \frac{dy}{dx} + y \log x = x e^x x^{-\frac{1}{2} \log x}$.
Dividing by $x$,we get $\frac{dy}{dx} + \left( \frac{\log x}{x} \right) y = e^x x^{-\frac{1}{2} \log x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$,where $P(x) = \frac{\log x}{x}$.
The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int \frac{\log x}{x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$. Thus,$\int \frac{\log x}{x} dx = \int u du = \frac{u^2}{2} = \frac{(\log x)^2}{2}$.
Therefore,$I.F. = e^{\frac{1}{2} (\log x)^2} = (e^{\frac{1}{2} \log x})^{\log x} = (e^{\log \sqrt{x}})^{\log x} = (\sqrt{x})^{\log x}$.

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = \sin x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \sin x$.
First,we find the integrating factor $(I.F.)$:
$I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values,we get $y \cdot x = \int x \sin x dx + c$.
Using integration by parts for $\int x \sin x dx$:
Let $u = x$ and $dv = \sin x dx$. Then $du = dx$ and $v = -\cos x$.
$\int x \sin x dx = -x \cos x - \int (-\cos x) dx = -x \cos x + \sin x$.
Thus,$xy = -x \cos x + \sin x + c$.
Rearranging the terms,we get $xy + x \cos x = \sin x + c$,which simplifies to $x(y + \cos x) = \sin x + c$.

(B) Given equation: $(x + \log y)dy + y\,dx = 0$
Rearranging the terms: $y\,dx + x\,dy + \log y\,dy = 0$
This can be written as: $d(xy) + \log y\,dy = 0$
Integrating both sides: $\int d(xy) + \int \log y\,dy = \int 0\,dc$
$xy + (y\log y - y) = c$
Thus,the solution is $xy + y\log y - y = c$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{1}{x + y + 1}$.
Taking the reciprocal,we get $\frac{dx}{dy} = x + y + 1$.
Rearranging the terms,we have $\frac{dx}{dy} - x = y + 1$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -1$ and $Q(y) = y + 1$.
The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int -1 dy} = e^{-y}$.
The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
$x e^{-y} = \int (y + 1) e^{-y} dy + c$.
Using integration by parts for $\int (y + 1) e^{-y} dy$:
Let $u = y + 1$ and $dv = e^{-y} dy$,then $du = dy$ and $v = -e^{-y}$.
$\int (y + 1) e^{-y} dy = (y + 1)(-e^{-y}) - \int (-e^{-y}) dy = -(y + 1)e^{-y} - e^{-y} + c = -(y + 2)e^{-y} + c$.
Thus,$x e^{-y} = -(y + 2)e^{-y} + c$.
Multiplying both sides by $e^y$,we get $x = -(y + 2) + ce^y$,which simplifies to $x = ce^y - y - 2$.

(A) Given differential equation is $\frac{dy}{dx} + 2xy = y$.
Rearranging the terms,we get $\frac{dy}{dx} + (2x - 1)y = 0$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2x - 1$ and $Q = 0$.
The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int (2x - 1) dx} = e^{x^2 - x}$.
The general solution is $y \times (I.F.) = \int (Q \times I.F.) dx + c$.
Substituting the values,$y \times e^{x^2 - x} = \int (0 \times e^{x^2 - x}) dx + c$.
$y \times e^{x^2 - x} = 0 + c$.
$y = c e^{-(x^2 - x)} = c e^{x - x^2}$.
Thus,the correct option is $A$.

(D) Given the differential equation: $x(1 - x^2)dy + (2x^2y - y - ax^3)dx = 0$.
Divide the entire equation by $x(1 - x^2)dx$:
$\frac{dy}{dx} + \frac{2x^2y - y - ax^3}{x(1 - x^2)} = 0$.
Rearranging the terms to fit the standard linear form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{(2x^2 - 1)y - ax^3}{x(1 - x^2)} = 0$.
$\frac{dy}{dx} + \frac{2x^2 - 1}{x(1 - x^2)}y = \frac{ax^3}{x(1 - x^2)}$.
Comparing this with the standard form $\frac{dy}{dx} + Py = Q$,we identify $P = \frac{2x^2 - 1}{x(1 - x^2)}$.

(D) Given the differential equation: $\cos x \, dy = y(\sin x - y) \, dx$
Divide both sides by $\cos x \, dx$:
$\frac{dy}{dx} = y \tan x - y^2 \sec x$
Rearrange the terms:
$\frac{dy}{dx} - y \tan x = -y^2 \sec x$
Divide by $y^2$:
$y^{-2} \frac{dy}{dx} - y^{-1} \tan x = -\sec x \quad \dots(1)$
Let $v = y^{-1} = \frac{1}{y}$. Then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,or $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.
Substituting into equation $(1)$:
$-\frac{dv}{dx} - v \tan x = -\sec x$
$\frac{dv}{dx} + v \tan x = \sec x$
This is a linear differential equation of the form $\frac{dv}{dx} + Pv = Q$,where $P = \tan x$ and $Q = \sec x$.
The Integrating Factor ($I$.$F$.) is $e^{\int P \, dx} = e^{\int \tan x \, dx} = e^{\ln|\sec x|} = \sec x$.
The general solution is $v \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c$.
$v \sec x = \int \sec x \cdot \sec x \, dx + c$
$v \sec x = \int \sec^2 x \, dx + c$
$v \sec x = \tan x + c$
Since $v = \frac{1}{y}$,we have:
$\frac{1}{y} \sec x = \tan x + c$
$\sec x = y(\tan x + c)$

(C) Given the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$.
Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.
The Integrating Factor $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
$x e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C$.
Let $t = -\frac{1}{y}$,then $dt = \frac{1}{y^2} dy$. Also,$\frac{1}{y} = -t$.
Substituting these,$x e^{-\frac{1}{y}} = \int (-t) e^t dt + C = - (t e^t - e^t) + C = (1 - t) e^t + C$.
$x e^{-\frac{1}{y}} = (1 + \frac{1}{y}) e^{-\frac{1}{y}} + C$.
Using $y(1) = 1$,we have $1 \cdot e^{-1} = (1 + 1) e^{-1} + C \Rightarrow e^{-1} = 2e^{-1} + C \Rightarrow C = -e^{-1} = -\frac{1}{e}$.
Thus,$x e^{-\frac{1}{y}} = (1 + \frac{1}{y}) e^{-\frac{1}{y}} - \frac{1}{e}$.
Dividing by $e^{-\frac{1}{y}}$,we get $x = 1 + \frac{1}{y} - \frac{e^{\frac{1}{y}}}{e}$.

(D) The given differential equation is $(x \log x) \frac{dy}{dx} + y = 2x \log x$.
Dividing by $(x \log x)$,we get $\frac{dy}{dx} + \frac{1}{x \log x} y = 2$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x \log x}$ and $Q(x) = 2$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.
The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.
$y \cdot \log x = \int 2 \log x dx = 2(x \log x - x) + C$.
Given $y(1) = 0$,substituting $x = 1$ gives $0 \cdot \log(1) = 2(1 \log 1 - 1) + C \Rightarrow 0 = 2(0 - 1) + C \Rightarrow C = 2$.
Thus,$y \log x = 2x \log x - 2x + 2$.
At $x = e$,$y \log e = 2e \log e - 2e + 2$.
Since $\log e = 1$,we have $y(1) = 2e - 2e + 2$,which simplifies to $y = 2$.

(A) The given differential equation is $(1 + t)\frac{dy}{dt} - ty = 1$.
Dividing by $(1 + t)$,we get $\frac{dy}{dt} - \frac{t}{1 + t}y = \frac{1}{1 + t}$.
This is a linear differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$,where $P(t) = -\frac{t}{1 + t}$ and $Q(t) = \frac{1}{1 + t}$.
The integrating factor ($I$.$F$.) is $e^{\int P(t) dt} = e^{\int -\frac{t}{1 + t} dt} = e^{\int (-1 + \frac{1}{1 + t}) dt} = e^{-t + \ln(1 + t)} = (1 + t)e^{-t}$.
The general solution is $y \cdot (I.F.) = \int Q(t) \cdot (I.F.) dt + C$.
$y(1 + t)e^{-t} = \int \frac{1}{1 + t} \cdot (1 + t)e^{-t} dt + C = \int e^{-t} dt + C = -e^{-t} + C$.
Given $y(0) = -1$,we substitute $t = 0$: $-1(1 + 0)e^{0} = -e^{0} + C \Rightarrow -1 = -1 + C \Rightarrow C = 0$.
Thus,$y(1 + t)e^{-t} = -e^{-t}$,which simplifies to $y = -\frac{1}{1 + t}$.
For $t = 1$,$y(1) = -\frac{1}{1 + 1} = -\frac{1}{2}$.

(A) Given equation: $ydx - xdy + \log x dx = 0$.
Dividing by $x dx$ (assuming $x \neq 0$),we get: $\frac{y}{x} - \frac{dy}{dx} + \frac{\log x}{x} = 0$.
Rearranging the terms: $\frac{dy}{dx} - \frac{y}{x} = \frac{\log x}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{1}{x}$ and $Q = \frac{\log x}{x}$.
The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$.
The solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values: $y \cdot \frac{1}{x} = \int \frac{\log x}{x} \cdot \frac{1}{x} dx + c = \int \frac{\log x}{x^2} dx + c$.
Using integration by parts for $\int \frac{\log x}{x^2} dx$: Let $u = \log x$ and $dv = x^{-2} dx$. Then $du = \frac{1}{x} dx$ and $v = -\frac{1}{x}$.
$\int \frac{\log x}{x^2} dx = (\log x)(-\frac{1}{x}) - \int (-\frac{1}{x})(\frac{1}{x}) dx = -\frac{\log x}{x} + \int x^{-2} dx = -\frac{\log x}{x} - \frac{1}{x}$.
Thus,$\frac{y}{x} = -\frac{\log x}{x} - \frac{1}{x} + c$.
Multiplying by $x$,we get $y = -\log x - 1 + cx$,which is $y = cx - (1 + \log x)$.

(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 - x^2}$ and $Q = \frac{x}{\sqrt{1 - x^2}}$.
The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx}$.
$I.F. = e^{\int \frac{2x}{1 - x^2} dx}$.
Let $u = 1 - x^2$,then $du = -2x dx$,so $2x dx = -du$.
$I.F. = e^{\int \frac{-du}{u}} = e^{-\ln|u|} = e^{-\ln|1 - x^2|} = e^{\ln|(1 - x^2)^{-1}|}$.
$I.F. = (1 - x^2)^{-1}$.

(D) The given differential equation is $y' - y \tan x = -2 \sin x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = -2 \sin x$.
The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P \, dx} = e^{-\int \tan x \, dx} = e^{-\ln |\sec x|} = e^{\ln |\cos x|} = \cos x$.
The solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c$.
Substituting the values,we get $y \cos x = \int (-2 \sin x \cos x) \, dx + c$.
$y \cos x = -\int \sin 2x \, dx + c$.
$y \cos x = \frac{\cos 2x}{2} + c$.
Multiplying by $\sec x$,we get $y = \frac{\cos 2x}{2 \cos x} + c \sec x$.
Since none of the given options match this result,the correct option is $D$.

(B) Given the differential equation: $(1 + y^2) + (x - e^{\tan^{-1}y}) \frac{dy}{dx} = 0$.
Rearranging the equation to solve for $\frac{dx}{dy}$:
$(1 + y^2) \frac{dx}{dy} + x = e^{\tan^{-1}y}$.
Dividing by $(1 + y^2)$:
$\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{\tan^{-1}y}}{1 + y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1 + y^2}$ and $Q(y) = \frac{e^{\tan^{-1}y}}{1 + y^2}$.
The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1}y}$.
The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + k$.
$x e^{\tan^{-1}y} = \int \frac{e^{\tan^{-1}y}}{1 + y^2} \cdot e^{\tan^{-1}y} dy + k$.
Let $u = \tan^{-1}y$,then $du = \frac{1}{1 + y^2} dy$.
$x e^{\tan^{-1}y} = \int e^{2u} du + k = \frac{e^{2u}}{2} + k = \frac{e^{2\tan^{-1}y}}{2} + k$.
Multiplying by $2$:
$2xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + k$.

(A) The given differential equation is $\frac{dy}{dx} + y\phi'(x) = \phi(x)\phi'(x)$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \phi'(x)$ and $Q = \phi(x)\phi'(x)$.
The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int \phi'(x) dx} = e^{\phi(x)}$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
$y \cdot e^{\phi(x)} = \int \phi(x)\phi'(x) e^{\phi(x)} dx + c$.
Let $t = \phi(x)$,then $dt = \phi'(x) dx$.
The integral becomes $\int t e^t dt = t e^t - e^t + c$.
Substituting back,$y \cdot e^{\phi(x)} = \phi(x)e^{\phi(x)} - e^{\phi(x)} + c$.
Dividing both sides by $e^{\phi(x)}$,we get $y = \phi(x) - 1 + c e^{-\phi(x)}$.