Linear differential equations Questions in English

(B) Given the differential equation: $(2x - 10y^3) dy + y dx = 0$
Rearranging the terms to form a linear differential equation in $x$:
$y dx = (10y^3 - 2x) dy$
$\frac{dx}{dy} = 10y^2 - \frac{2x}{y}$
$\frac{dx}{dy} + \frac{2}{y} x = 10y^2$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{2}{y}$ and $Q = 10y^2$.
The integrating factor $(IF)$ is given by:
$IF = e^{\int P dy} = e^{\int \frac{2}{y} dy} = e^{2 \ln y} = y^2$
The general solution is $x \cdot (IF) = \int Q \cdot (IF) dy + C$:
$x \cdot y^2 = \int (10y^2) \cdot y^2 dy + C$
$x y^2 = \int 10y^4 dy + C$
$x y^2 = 10 \cdot \frac{y^5}{5} + C$
$x y^2 = 2y^5 + C$
Thus,the general solution is $x y^2 - 2y^5 = C$.

(D) Given the differential equation: $(\sec x + \tan x) \frac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1$
Divide by $(\sec x + \tan x)$:
$\frac{dy}{dx} + \sec x \cdot y = \frac{1}{\sec x + \tan x}$
Since $\frac{1}{\sec x + \tan x} = \sec x - \tan x$,the equation becomes:
$\frac{dy}{dx} + (\sec x) y = \sec x - \tan x$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \sec x$ and $Q = \sec x - \tan x$.
Integrating Factor $(IF)$ $= e^{\int P dx} = e^{\int \sec x dx} = e^{\ln|\sec x + \tan x|} = \sec x + \tan x$.
The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + c$.
$y(\sec x + \tan x) = \int (\sec x - \tan x)(\sec x + \tan x) dx + c$
$y(\sec x + \tan x) = \int (\sec^2 x - \tan^2 x) dx + c$
Since $\sec^2 x - \tan^2 x = 1$:
$y(\sec x + \tan x) = \int 1 dx + c$
$y(\sec x + \tan x) = x + c$.

(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^2$.
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 $(IF)$:
$IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$.
Substituting the values,we get $y \cdot x = \int (x^2 \cdot x) dx + C$.
$xy = \int x^3 dx + C$.
Integrating $x^3$,we get $xy = \frac{x^4}{4} + C$.

(A) Given the differential equation: $x \frac{dy}{dx} = x^2 + 3y$
Divide by $x$ $(x > 0)$: $\frac{dy}{dx} - \frac{3}{x}y = x$
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{3}{x}$ and $Q(x) = x$.
The Integrating Factor $(IF)$ is: $IF = e^{\int P(x) dx} = e^{\int -\frac{3}{x} dx} = e^{-3 \ln x} = e^{\ln x^{-3}} = x^{-3} = \frac{1}{x^3}$.
The general solution is: $y \cdot IF = \int Q(x) \cdot IF dx + C$
$y \cdot \frac{1}{x^3} = \int x \cdot \frac{1}{x^3} dx + C$
$\frac{y}{x^3} = \int x^{-2} dx + C = -x^{-1} + C = -\frac{1}{x} + C$
Given $y(2) = 4$,substitute $x = 2$ and $y = 4$:
$\frac{4}{2^3} = -\frac{1}{2} + C \Rightarrow \frac{4}{8} = -\frac{1}{2} + C \Rightarrow \frac{1}{2} = -\frac{1}{2} + C \Rightarrow C = 1$.
Thus,the particular solution is: $\frac{y}{x^3} = -\frac{1}{x} + 1 \Rightarrow y = x^3 - x^2$.
To find $f(4)$,substitute $x = 4$:
$f(4) = 4^3 - 4^2 = 64 - 16 = 48$.

(C) We know that the length of the sub-tangent is given by $\frac{y}{dy/dx}$.
Given,$\frac{y}{dy/dx} = x + 7y^2$.
Rearranging the terms,we get $\frac{dy}{dx} = \frac{y}{x + 7y^2}$.
Taking the reciprocal,we have $\frac{dx}{dy} = \frac{x + 7y^2}{y} = \frac{x}{y} + 7y$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 7y$.
The integrating factor $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$.
The solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
$x \cdot \frac{1}{y} = \int 7y \cdot \frac{1}{y} dy + C$.
$\frac{x}{y} = \int 7 dy + C = 7y + C$.
$x = 7y^2 + Cy$.
Thus,$7y^2 + Cy - x = 0$,which corresponds to $f(x, y) = 7y^2 + cy - x$.

(A) Given the condition $x f^{\prime}(x) = 2 f(x) - 2 x^3$ for $x \in (2, 5)$.
Rearranging the terms,we get the linear differential equation: $\frac{d y}{d x} - \frac{2}{x} y = -2 x^2$,where $y = f(x)$.
The integrating factor is $IF = e^{\int -\frac{2}{x} d x} = e^{-2 \ln x} = \frac{1}{x^2}$.
Multiplying both sides by the integrating factor: $\frac{d}{d x} \left( \frac{y}{x^2} \right) = -2$.
Integrating both sides with respect to $x$: $\frac{y}{x^2} = -2 x + C$,which gives $f(x) = -2 x^3 + C x^2$.
Given $\frac{f(5)}{f(2)} = 1$,we have $f(5) = f(2)$.
Substituting $x = 5$ and $x = 2$: $-2(125) + C(25) = -2(8) + C(4)$.
$-250 + 25 C = -16 + 4 C$.
$21 C = 234 \Rightarrow C = \frac{234}{21} = \frac{78}{7}$.
Thus,$f(x) = -2 x^3 + \frac{78}{7} x^2$.

(A) Given differential equation is $(1+y^2) dx = (\tan^{-1} y - x) dy$.
Dividing by $(1+y^2) dy$,we get $\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2}$.
Rearranging the terms,we have $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1} y}{1+y^2}$.
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 $Q(y) = \frac{\tan^{-1} y}{1+y^2}$.
The integrating factor $IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.
The general solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + C$.
$x \cdot e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + C$.
Let $t = \tan^{-1} y$,then $dt = \frac{1}{1+y^2} dy$.
$x \cdot e^{\tan^{-1} y} = \int t e^t dt + C$.
Using integration by parts,$\int t e^t dt = t e^t - e^t$.
$x \cdot e^{\tan^{-1} y} = e^t(t - 1) + C$.
Substituting $t = \tan^{-1} y$,we get $x \cdot e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + C$.
Dividing by $e^{\tan^{-1} y}$,we get $x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}$.

(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = e^{2x}$.
Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{e^{2x}}{\sin x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \cot x$ and $Q(x) = \frac{e^{2x}}{\sin x}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x$.
Multiplying the equation by the $I$.$F$.,we get $\frac{d}{dx}(y \sin x) = e^{2x}$.
Integrating both sides with respect to $x$,we get $y \sin x = \int e^{2x} dx = \frac{e^{2x}}{2} + C$.
Thus,$y = \frac{e^{2x}}{2 \sin x} + \frac{C}{\sin x}$.
Using the condition $y\left(\frac{\pi}{2}\right) = 0$,we have $0 = \frac{e^\pi}{2(1)} + \frac{C}{1}$,which gives $C = -\frac{e^\pi}{2}$.
Substituting $C$ back,$y = \frac{e^{2x} - e^\pi}{2 \sin x}$.
Now,for $x = \frac{\pi}{6}$,$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.
Therefore,$y\left(\frac{\pi}{6}\right) = \frac{e^{\pi/3} - e^\pi}{2(1/2)} = e^{\pi/3} - e^\pi$.

(D) Given the differential equation: $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$.
Dividing by $x \cos x$,we get: $\frac{dy}{dx} + (\tan x + \frac{1}{x}) y = \frac{\sec x}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x + \frac{1}{x}$ and $Q = \frac{\sec x}{x}$.
The Integrating Factor $(IF)$ is given by: $IF = e^{\int P dx} = e^{\int (\tan x + \frac{1}{x}) dx} = e^{\ln(\sec x) + \ln(x)} = e^{\ln(x \sec x)} = x \sec x$.
The solution is given by: $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
Substituting the values: $y(x \sec x) = \int \frac{\sec x}{x} \cdot (x \sec x) dx + C$.
$x y \sec x = \int \sec^2 x dx + C$.
$x y \sec x = \tan x + C$.
Thus,$x y \sec x - \tan x = C$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{1}{ax + 4y + 7}$.
Taking the reciprocal,we get: $\frac{dx}{dy} = ax + 4y + 7$.
Rearranging the terms,we have: $\frac{dx}{dy} - ax = 4y + 7$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -a$ and $Q = 4y + 7$.
Since the equation can be written in the form $\frac{dx}{dy} + Px = Q$,it is linear in $x$.
Because the equation is linear in $x$,it is not linear in $y$.
Thus,statement $A$ is true and statement $B$ is true.
The integrating factor $(IF)$ is given by $e^{\int P dy} = e^{\int -a dy} = e^{-ay}$.
Therefore,statement $D$ is false,and statement $C$ is false.
Hence,only $A$ and $B$ are true.

(A) Given differential equation is $\sqrt{1-y^2} dx + (x - \sin^{-1} y) dy = 0$.
Dividing by $dy$ and $\sqrt{1-y^2}$,we get $\frac{dx}{dy} + \frac{x}{\sqrt{1-y^2}} = \frac{\sin^{-1} y}{\sqrt{1-y^2}}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{\sqrt{1-y^2}}$ and $Q(y) = \frac{\sin^{-1} y}{\sqrt{1-y^2}}$.
The integrating factor is $IF = e^{\int P(y) dy} = e^{\int \frac{1}{\sqrt{1-y^2}} dy} = e^{\sin^{-1} y}$.
The solution is $x \cdot IF = \int Q(y) \cdot IF dy + c$.
$x e^{\sin^{-1} y} = \int \frac{\sin^{-1} y}{\sqrt{1-y^2}} e^{\sin^{-1} y} dy + c$.
Let $t = \sin^{-1} y$,then $dt = \frac{1}{\sqrt{1-y^2}} dy$.
$x e^{\sin^{-1} y} = \int t e^t dt + c = (t e^t - e^t) + c = e^t(t - 1) + c$.
Substituting $t = \sin^{-1} y$,we get $x e^{\sin^{-1} y} = e^{\sin^{-1} y}(\sin^{-1} y - 1) + c$.
Dividing by $e^{\sin^{-1} y}$,we get $x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}$.

(C) Given the differential equation: $(1+y^2)+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$.
Rearranging the equation: $(x-e^{\tan ^{-1} y}) \frac{dy}{dx} = -(1+y^2)$.
Dividing by $\frac{dy}{dx}$,we get $\frac{dx}{dy} = -\frac{x-e^{\tan ^{-1} y}}{1+y^2} = -\frac{x}{1+y^2} + \frac{e^{\tan ^{-1} y}}{1+y^2}$.
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 general solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
$x e^{\tan ^{-1} y} = \int \frac{e^{\tan ^{-1} y}}{1+y^2} \cdot e^{\tan ^{-1} y} dy + c = \int \frac{e^{2 \tan ^{-1} y}}{1+y^2} dy + c$.
Let $u = \tan ^{-1} y$,then $du = \frac{1}{1+y^2} dy$.
$x e^{\tan ^{-1} y} = \int e^{2u} du + c = \frac{1}{2} e^{2u} + c = \frac{1}{2} e^{2 \tan ^{-1} y} + c$.
Multiplying by $2$,we get $2x e^{\tan ^{-1} y} = e^{2 \tan ^{-1} y} + 2c$.
Since $2c$ is a constant,we can write the solution as $2x e^{\tan ^{-1} y} - e^{2 \tan ^{-1} y} = C$.

(B) Given the differential equation: $(x+2y^3) \frac{dy}{dx} = y$
Rearranging the equation,we get: $y \frac{dx}{dy} = x + 2y^3$
Dividing by $y$,we get: $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$
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 $(IF)$ is given by: $IF = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.
The solution is given by: $x \cdot (IF) = \int (Q \cdot IF) dy + c$
Substituting the values: $x \cdot \frac{1}{y} = \int (2y^2 \cdot \frac{1}{y}) dy + c$
$\frac{x}{y} = \int 2y dy + c$
$\frac{x}{y} = y^2 + c$
Therefore,$x = y^3 + cy$.

(D) Given the differential equation: $(1+y^2) dx = ( an^{-1} y - x) dy$
Rearranging the terms,we get: $\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2}$
$\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1} y}{1+y^2}$
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 $(IF)$ is given by $IF = e^{\int P dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.
The general solution is $x \cdot IF = \int (Q \cdot IF) dy + c$.
$x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + c$.
Let $t = \tan^{-1} y$,then $dt = \frac{1}{1+y^2} dy$.
$x e^{\tan^{-1} y} = \int t e^t dt + c$.
Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1)$.
Substituting back $t = \tan^{-1} y$,we get: $x e^{\tan^{-1} y} = e^{\tan^{-1} y} (\tan^{-1} y - 1) + c$.

(B) Given differential equation is: $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$
Dividing by $\left(\frac{1+x^3}{x^2}\right)$,we get: $\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{x^2}{1+x^3}$
This is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$,where $P=\frac{3 x^2}{1+x^3}$ and $Q=\frac{x^2}{1+x^3}$.
The Integrating Factor $(IF)$ is given by: $IF = e^{\int P d x} = e^{\int \frac{3 x^2}{1+x^3} d x} = e^{\log(1+x^3)} = 1+x^3$.
The general solution is $y \cdot IF = \int (Q \cdot IF) d x + c$.
Substituting the values: $y(1+x^3) = \int \left(\frac{x^2}{1+x^3}\right)(1+x^3) d x + c$.
$y(1+x^3) = \int x^2 d x + c$.
$y(1+x^3) = \frac{x^3}{3} + c$.
Multiplying by $3$: $3y(1+x^3) = x^3 + 3c$.
$3y + 3yx^3 - x^3 = 3c$.
$(3y-1)x^3 + 3y = C$ (where $C = 3c$).

(D) Given the differential equation:
$(1+y^2) dx = (\tan^{-1} y - x) dy$
Rearranging the terms to the form $\frac{dx}{dy} + P(y)x = Q(y)$:
$\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2}$
$\frac{dx}{dy} + \frac{1}{1+y^2} x = \frac{\tan^{-1} y}{1+y^2}$
This is a linear differential equation in $x$ where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan^{-1} y}{1+y^2}$.
The Integrating Factor $(IF)$ is:
$IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$
The general solution is given by:
$x \cdot IF = \int Q(y) \cdot IF dy + k$
$x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + k$
Let $t = \tan^{-1} y$,then $dt = \frac{1}{1+y^2} dy$:
$x e^{\tan^{-1} y} = \int t e^t dt + k$
Using integration by parts ($\int u dv = uv - \int v du$ with $u=t, dv=e^t dt$):
$x e^{\tan^{-1} y} = t e^t - e^t + k$
$x e^{\tan^{-1} y} = e^{\tan^{-1} y} (\tan^{-1} y - 1) + k$
Dividing by $e^{\tan^{-1} y}$:
$x = (\tan^{-1} y - 1) + k e^{-\tan^{-1} y}$

(A) Given differential equation is $\frac{dy}{dx} = \frac{1}{e^y - x}$.
Taking the reciprocal on both sides,we get:
$\frac{dx}{dy} = e^y - x$.
Rearranging the terms,we get the linear differential equation in $x$:
$\frac{dx}{dy} + x = e^y$.
Comparing this with the standard form $\frac{dx}{dy} + Px = Q$,where $P = 1$ and $Q = e^y$.
The integrating factor $(IF)$ is given by:
$IF = e^{\int P dy} = e^{\int 1 dy} = e^y$.
The solution is given by $x \cdot (IF) = \int (Q \cdot IF) dy + C$.
$x \cdot e^y = \int (e^y \cdot e^y) dy + C$.
$x \cdot e^y = \int e^{2y} dy + C$.
$x \cdot e^y = \frac{e^{2y}}{2} + C$.
Dividing by $e^y$,we get $x = \frac{e^y}{2} + Ce^{-y}$.
Wait,re-evaluating the original equation $y' = \frac{1}{e^y - x}$ as $\frac{dx}{dy} = e^y - x$.
If the equation was $y' = \frac{1}{e^{-y} - x}$,then $\frac{dx}{dy} = e^{-y} - x \Rightarrow \frac{dx}{dy} + x = e^{-y}$.
$IF$ $= e^{\int 1 dy} = e^y$.
$x \cdot e^y = \int e^{-y} \cdot e^y dy + C = \int 1 dy + C = y + C$.
$x = e^{-y}(y + C)$.
Thus,the correct option is $A$.

(A) The given differential equation is $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$.
Rearranging the terms,we get $(x \sin y - 1) \frac{dy}{dx} = -\cos y$.
Taking the reciprocal,we have $\frac{dx}{dy} = -\frac{x \sin y - 1}{\cos y} = -x \tan y + \sec y$.
This can be written as $\frac{dx}{dy} + x \tan y = \sec y$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \tan y$ and $Q = \sec y$.
The integrating factor $(IF)$ is $e^{\int P dy} = e^{\int \tan y dy} = e^{\ln |\sec y|} = \sec y$.
The general solution is given by $x(IF) = \int Q(IF) dy + C$.
Substituting the values,$x \sec y = \int \sec y \cdot \sec y dy + C$.
$x \sec y = \int \sec^2 y dy + C$.
$x \sec y = \tan y + C$.

(D) The given differential equation is $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$.
Dividing both sides by $(1-x^2)$,we get:
$\frac{d y}{d x} + \frac{x}{1-x^2} y = \frac{x^4 (1-x^2)^{3/2}}{(1+x^5)(1-x^2)} = \frac{x^4 \sqrt{1-x^2}}{1+x^5}$.
This is a linear differential equation of the form $\frac{d y}{d x} + P(x) y = Q(x)$,where $P(x) = \frac{x}{1-x^2}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int \frac{x}{1-x^2} dx}$.
Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
$IF = e^{-\frac{1}{2} \int \frac{1}{u} du} = e^{-\frac{1}{2} \ln|u|} = e^{\ln|u|^{-1/2}} = |u|^{-1/2} = \frac{1}{\sqrt{1-x^2}}$.

(B) Given,$(x+y+1) \frac{dy}{dx} = 1$
$\Rightarrow \frac{dx}{dy} = x+y+1$
$\Rightarrow \frac{dx}{dy} - x = y+1$,which is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$.
Here,$P = -1$ and $Q = y+1$.
The Integrating Factor $(IF)$ is $e^{\int P dy} = e^{\int -1 dy} = e^{-y}$.
The solution is given by $x \cdot (IF) = \int Q \cdot (IF) dy + c$.
$x e^{-y} = \int (y+1) e^{-y} dy + c$
Using integration by parts for $\int y e^{-y} dy$:
$x e^{-y} = [y(-e^{-y}) - \int 1 \cdot (-e^{-y}) dy] + \int e^{-y} dy + c$
$x e^{-y} = -y e^{-y} - e^{-y} - e^{-y} + c$
$x e^{-y} = -(y+2) e^{-y} + c$
Multiplying both sides by $e^y$,we get:
$x = -(y+2) + ce^y$.

(B) Given the differential equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$
Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$
Let $t = \frac{1}{x}$,then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$
Substituting this into the equation: $-\frac{dt}{dy} + \frac{t}{y} = 1$
Rearranging gives: $\frac{dt}{dy} - \frac{t}{y} = -1$
This is a linear differential equation of the form $\frac{dt}{dy} + P(y)t = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -1$
Integrating Factor ($I$.$F$.) $= e^{\int P(y) dy} = e^{-\int \frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$
The general solution is $t \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$
$t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$
$\frac{1}{xy} = -\log y + c$
Multiplying by $y$,we get: $\frac{1}{x} = cy - y \log y$

(B) Given the linear differential equation: $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$.
Dividing by $(1+x^2)$,we get: $\frac{dy}{dx} + \left(\frac{2x}{1+x^2}\right)y = \frac{4x^2}{1+x^2}$.
This is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1+x^2}$ and $Q = \frac{4x^2}{1+x^2}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values: $y(1+x^2) = \int \left(\frac{4x^2}{1+x^2}\right)(1+x^2) dx + c$.
$y(1+x^2) = \int 4x^2 dx + c$.
$y(1+x^2) = \frac{4x^3}{3} + c$.
Multiplying by $3$,we get: $3y(1+x^2) = 4x^3 + c$.

(C) For statement $A$:
The given differential equation is $\frac{dy}{dx} + y = x^2$.
Comparing this with the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$,we get $P(x) = 1$.
The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 1 dx} = e^x$.
Thus,statement $A$ is true.
For statement $R$:
The standard form of a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.
The integrating factor is defined as $e^{\int P(x) dx}$.
Thus,statement $R$ is true.
Since statement $A$ is derived directly from the formula given in statement $R$,$R \Rightarrow A$ is true.
Therefore,both statements are true and $R$ is the correct explanation for $A$.

(A) The given differential equation is $\left(x+2 y^3\right) \frac{d y}{d x}=y^2$.
Rearranging the equation to the form $\frac{dx}{dy} + P(y)x = Q(y)$,we get:
$\frac{dx}{dy} = \frac{x+2y^3}{y^2} = \frac{x}{y^2} + 2y$
$\frac{dx}{dy} - \frac{1}{y^2}x = 2y$
Here,$P(y) = -\frac{1}{y^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy}$.
$IF = e^{\int -\frac{1}{y^2} dy} = e^{\int -y^{-2} dy} = e^{-(-y^{-1})} = e^{\frac{1}{y}}$.

(C) The given differential equation is $\frac{dy}{dx} + y = e^x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^x$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
Substituting the values,we get $y \cdot e^x = \int e^x \cdot e^x dx + C$.
$y e^x = \int e^{2x} dx + C$.
$y e^x = \frac{e^{2x}}{2} + C$.
Multiplying both sides by $2$,we get $2ye^x = e^{2x} + 2C$.
Letting $2C = C_1$,we have $2ye^x = e^{2x} + C_1$.

(C) Given the differential equation: $(y-3 x^2) d x+x d y=0$.
Rearranging the terms,we get: $y d x-3 x^2 d x+x d y=0$.
Grouping the terms $y d x$ and $x d y$,we have: $y d x+x d y=3 x^2 d x$.
Recognizing the product rule for differentiation,$d(x y) = y d x + x d y$,the equation becomes: $d(x y) = 3 x^2 d x$.
Integrating both sides: $\int d(x y) = \int 3 x^2 d x$.
This yields: $x y = x^3 + C$.
Dividing by $x$ (assuming $x \neq 0$),we get the solution: $y = x^2 + \frac{C}{x}$.

(D) Given differential equation: $x \frac{d y}{d x} - y = x^2 \sin x$.
Dividing by $x^2$,we get $\frac{x \frac{d y}{d x} - y}{x^2} = \sin x$.
This is the derivative of $\frac{y}{x}$,so $\frac{d}{d x} \left( \frac{y}{x} \right) = \sin x$.
Integrating both sides: $\frac{y}{x} = -\cos x + c$.
Given $y(\pi) = 0$,we have $\frac{0}{\pi} = -\cos(\pi) + c \Rightarrow 0 = 1 + c \Rightarrow c = -1$.
Thus,$y = -x \cos x - x$.
For an extreme value at $x = \alpha$,we set $\frac{d y}{d x} = 0$.
$\frac{d y}{d x} = -(\cos x - x \sin x) - 1 = -\cos x + x \sin x - 1 = 0$.
At $x = \alpha$,$x \sin x - \cos x - 1 = 0$.
Using trigonometric identities: $\alpha \sin \alpha - (\cos \alpha + 1) = 0$.
$\alpha (2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}) - (2 \cos^2 \frac{\alpha}{2}) = 0$.
$2 \cos \frac{\alpha}{2} (\alpha \sin \frac{\alpha}{2} - \cos \frac{\alpha}{2}) = 0$.
Since $\cos \frac{\alpha}{2} \neq 0$ for the existence of the function,we have $\alpha \sin \frac{\alpha}{2} = \cos \frac{\alpha}{2}$,which implies $\alpha = \cot \frac{\alpha}{2}$.

(B) Differentiating the given equation $\int_0^x (f^{\prime}(t)-\sin 2t) dt = \int_x^0 f(t) \tan t dt$ with respect to $x$ using the Leibniz rule,we get:
$f^{\prime}(x) - \sin 2x = -f(x) \tan x$
$f^{\prime}(x) + f(x) \tan x = \sin 2x$
This is a linear differential equation of the form $\frac{df}{dx} + P(x)f = Q(x)$,where $P(x) = \tan x$ and $Q(x) = \sin 2x = 2 \sin x \cos x$.
The integrating factor $IF = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
Multiplying by $IF$,we get $\frac{d}{dx} (f(x) \sec x) = \sin 2x \sec x = 2 \sin x$.
Integrating both sides,$f(x) \sec x = \int 2 \sin x dx = -2 \cos x + C$.
Given $f(0) = 1$,we have $1 \cdot \sec 0 = -2 \cos 0 + C \implies 1 = -2 + C \implies C = 3$.
Thus,$f(x) \sec x = -2 \cos x + 3$,which implies $f(x) = -2 \cos^2 x + 3 \cos x$.
Now,$\int_0^{\frac{\pi}{2}} f(x) dx = \int_0^{\frac{\pi}{2}} (-2 \cos^2 x + 3 \cos x) dx$.
Using $\cos^2 x = \frac{1+\cos 2x}{2}$,we get $\int_0^{\frac{\pi}{2}} (-1 - \cos 2x + 3 \cos x) dx = [-x - \frac{\sin 2x}{2} + 3 \sin x]_0^{\frac{\pi}{2}}$.
$= (-\frac{\pi}{2} - 0 + 3) - (0) = 3 - \frac{\pi}{2}$.

(A) Given,$f(x) = \int_{0}^{x} f(t) \, dt$.
Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:
$f'(x) = f(x)$.
This is a first-order linear differential equation. Separating variables,we get $\frac{f'(x)}{f(x)} = 1$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = k e^{x}$ for some constant $k$.
Now,substitute $x = 0$ into the original equation:
$f(0) = \int_{0}^{0} f(t) \, dt = 0$.
Using $f(0) = k e^{0} = k$,we find $k = 0$.
Therefore,$f(x) = 0 \cdot e^{x} = 0$ for all $x \in R$.
Thus,$f(\log_{e} 5) = 0$.

(A, B, D) Let $f(x)=\cos x$ and $g(x)=\sin x$. Consider the Wronskian of $f(x)$ and $g(x)$.
$W = \begin{vmatrix} f(x) & g(x) \\ f'(x) & g'(x) \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1 \neq 0$.
Since the Wronskian is non-zero,the functions are linearly independent. The general solution is $y = A \cos x + B \sin x$.
$(a)$ $A \cos x + B \sin x$ is the general solution,so it is true.
$(b)$ $A \cos(x + \frac{\pi}{4}) = A(\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}) = \frac{A}{\sqrt{2}} \cos x - \frac{A}{\sqrt{2}} \sin x$. This is of the form $C_1 \cos x + C_2 \sin x$,so it is true.
$(c)$ $A \cos x \sin x = \frac{A}{2} \sin(2x)$,which is not of the form $A \cos x + B \sin x$,so it is false.
$(d)$ $A \cos(x + \frac{\pi}{4}) + B \sin(x - \frac{\pi}{4}) = A(\frac{\cos x - \sin x}{\sqrt{2}}) + B(\frac{\sin x - \cos x}{\sqrt{2}}) = \cos x(\frac{A-B}{\sqrt{2}}) + \sin x(\frac{B-A}{\sqrt{2}})$. This is of the form $C_1 \cos x + C_2 \sin x$,so it is true.

(A) The given differential equation $\frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0$ is a linear homogeneous differential equation of the second order.
By the principle of superposition,if $u(x)$ and $v(x)$ are two independent solutions,then any linear combination $y = c_{1}u(x) + c_{2}v(x)$ is also a solution,where $c_{1}$ and $c_{2}$ are arbitrary constants.
Option $A$ is a specific case of a linear combination where $c_{1}=5$ and $c_{2}=8$.
Option $B$ can be rewritten as $y = c_{1}u(x) + (c_{2}-c_{1})v(x)$,which is also a linear combination of $u(x)$ and $v(x)$ with new arbitrary constants.
Therefore,both $A$ and $B$ represent solutions to the differential equation.

(B) Given the differential equation: $y \frac{dy}{dx} + by^2 = a \cos x$.
Let $y^2 = z$. Then,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = \frac{dz}{dx}$,which implies $y \frac{dy}{dx} = \frac{1}{2} \frac{dz}{dx}$.
Substituting this into the original equation: $\frac{1}{2} \frac{dz}{dx} + bz = a \cos x$.
Multiplying by $2$: $\frac{dz}{dx} + 2bz = 2a \cos x$.
This is a linear differential equation of the form $\frac{dz}{dx} + Pz = Q$,where $P = 2b$ and $Q = 2a \cos x$.
The integrating factor $(IF)$ is $e^{\int 2b \, dx} = e^{2bx}$.
The solution is given by $z \cdot IF = \int Q \cdot IF \, dx + c$.
$z e^{2bx} = \int 2a \cos x \cdot e^{2bx} \, dx + c$.
Using the standard integral $\int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$,we get:
$z e^{2bx} = 2a \left[ \frac{e^{2bx}}{(2b)^2 + 1^2} (2b \cos x + \sin x) \right] + c$.
$y^2 e^{2bx} = \frac{2a}{4b^2 + 1} e^{2bx} (2b \cos x + \sin x) + c$.
Multiplying by $e^{-2bx}$: $y^2 = \frac{2a}{4b^2 + 1} (2b \cos x + \sin x) + c e^{-2bx}$.
Rearranging gives: $(4b^2 + 1) y^2 = 2a(\sin x + 2b \cos x) + c e^{-2bx}$.

(B) The slope of the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = y + 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} - y = 2x$.
The integrating factor is $IF = e^{\int -1 dx} = e^{-x}$.
Multiplying both sides by $e^{-x}$,we get $e^{-x} \frac{dy}{dx} - y e^{-x} = 2x e^{-x}$.
This can be written as $\frac{d}{dx}(y e^{-x}) = 2x e^{-x}$.
Integrating both sides with respect to $x$:
$y e^{-x} = \int 2x e^{-x} dx$.
Using integration by parts: $\int 2x e^{-x} dx = 2x(-e^{-x}) - \int 2(-e^{-x}) dx = -2x e^{-x} - 2e^{-x} + C$.
So,$y e^{-x} = -2x e^{-x} - 2e^{-x} + C$.
Multiplying by $e^x$,we get $y = -2x - 2 + C e^x$.
For a curve passing through the origin $(0,0)$,$0 = 0 - 2 + C(1) \Rightarrow C = 2$.
Thus,$y = 2e^x - 2x - 2 = 2(e^x - x - 1)$.

(C) The given differential equation is $25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$.
The auxiliary equation is $25 m^{2}-10 m+1=0$.
This can be written as $(5 m-1)^{2}=0$,which gives $m=\frac{1}{5}, \frac{1}{5}$.
Since the roots are real and equal,the general solution is $y=(c_{1}+c_{2} x) e^{x / 5} \quad \dots(i)$.
Given $y(0)=1$,substituting $x=0$ in $(i)$ gives $1=(c_{1}+0) e^{0} \Rightarrow c_{1}=1$.
Given $y(1)=2 e^{1 / 5}$,substituting $x=1$ and $c_{1}=1$ in $(i)$ gives $2 e^{1 / 5}=(1+c_{2}) e^{1 / 5}$.
Dividing by $e^{1 / 5}$,we get $2=1+c_{2} \Rightarrow c_{2}=1$.
Substituting $c_{1}=1$ and $c_{2}=1$ in $(i)$,the particular solution is $y=(1+x) e^{x / 5}$.

(B) The given differential equation is $\frac{dy}{dx} + y f'(x) = f(x) f'(x)$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = f'(x)$ and $Q(x) = f(x) f'(x)$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int f'(x) dx} = e^{f(x)}$.
The solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y e^{f(x)} = \int f(x) f'(x) e^{f(x)} dx + C$.
Let $t = f(x)$,then $dt = f'(x) dx$.
$y e^{f(x)} = \int t e^t dt + C = e^t(t-1) + C$.
Substituting back $t = f(x)$,we get $y e^{f(x)} = e^{f(x)}(f(x)-1) + C$.
Given $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$,we substitute these limits into the equation:
$0 \cdot e^0 = e^0(0-1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.
Thus,$y e^{f(x)} = e^{f(x)}(f(x)-1) + 1$.
Dividing by $e^{f(x)}$,we get $y = f(x) - 1 + e^{-f(x)}$.
Rearranging gives $y + 1 = f(x) + e^{-f(x)}$.

(C) The given differential equation is $(1+x^{2}) \frac{dy}{dx} + 2xy = 4x^{2}$.
Dividing by $(1+x^{2})$,we get $\frac{dy}{dx} + \left(\frac{2x}{1+x^{2}}\right)y = \frac{4x^{2}}{1+x^{2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{1+x^{2}}$ and $Q(x) = \frac{4x^{2}}{1+x^{2}}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^{2}} dx} = e^{\ln(1+x^{2})} = 1+x^{2}$.
The general solution is given by $y(IF) = \int Q(x)(IF) dx + C$.
$y(1+x^{2}) = \int \left(\frac{4x^{2}}{1+x^{2}}\right)(1+x^{2}) dx + C$.
$y(1+x^{2}) = \int 4x^{2} dx + C = \frac{4x^{3}}{3} + C$.
Using the initial condition $y(0) = -1$,we substitute $x=0$ and $y=-1$:
$-1(1+0^{2}) = \frac{4(0)^{3}}{3} + C \Rightarrow C = -1$.
Thus,$y(1+x^{2}) = \frac{4x^{3}}{3} - 1$.
For $x=1$,$y(1+1^{2}) = \frac{4(1)^{3}}{3} - 1$.
$2y = \frac{4}{3} - 1 = \frac{1}{3}$.
Therefore,$y(1) = \frac{1}{6}$.

(B) Given the differential equation: $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$.
Divide throughout by $x^{2}(x^{2}-1)$ to get it in the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{x^{2}+1}{x(x^{2}-1)}y = \frac{1}{x^{2}}$.
Here,$P = \frac{x^{2}+1}{x(x^{2}-1)}$.
The integrating factor $(IF)$ is given by $e^{\int P dx}$.
$\int P dx = \int \frac{x^{2}+1}{x(x-1)(x+1)} dx$.
Using partial fractions: $\frac{x^{2}+1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.
Solving for constants: $x^{2}+1 = A(x^{2}-1) + Bx(x+1) + Cx(x-1)$.
For $x=0$,$1 = -A \Rightarrow A = -1$.
For $x=1$,$2 = 2B \Rightarrow B = 1$.
For $x=-1$,$2 = 2C \Rightarrow C = 1$.
So,$\int P dx = \int (\frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) dx = -\ln|x| + \ln|x-1| + \ln|x+1| = \ln|\frac{x^{2}-1}{x}|$.
$IF = e^{\ln|\frac{x^{2}-1}{x}|} = \frac{x^{2}-1}{x} = x - \frac{1}{x}$.

(B) Given the differential equation: $x \frac{dy}{dx} + y = x e^x$.
Dividing by $x$ (assuming $x \neq 0$),we get: $\frac{dy}{dx} + \frac{1}{x} y = e^x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = e^x$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.
The solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
$xy = \int x e^x dx + C$.
Using integration by parts for $\int x e^x dx$:
$\int x e^x dx = x e^x - \int 1 \cdot e^x dx = x e^x - e^x = e^x(x-1)$.
Thus,$xy = e^x(x-1) + C$.
Comparing this with the given form $xy = e^x \phi(x) + C$,we find $\phi(x) = x-1$.

(B) Given the differential equation: $\frac{dy}{dx} = -(3x^2 \tan^{-1} y - x^3)(1 + y^2)$.
Rearranging the terms: $\frac{dy}{dx} = x^3(1 + y^2) - 3x^2(\tan^{-1} y)(1 + y^2)$.
Dividing both sides by $(1 + y^2)$: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} = x^3 - 3x^2 \tan^{-1} y$.
Rearranging to standard form: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} + 3x^2 \tan^{-1} y = x^3$.
Let $t = \tan^{-1} y$,then $\frac{dt}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx}$.
The equation becomes: $\frac{dt}{dx} + 3x^2 t = x^3$.
This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = 3x^2$.
The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 3x^2 dx} = e^{x^3}$.

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

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log_{e} x}$ and $Q = \frac{1}{x}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{x \log_{e} x} dx}$.
Let $u = \log_{e} x$,then $du = \frac{1}{x} dx$. Thus,$\int \frac{1}{x \log_{e} x} dx = \int \frac{1}{u} du = \log_{e} u = \log_{e}(\log_{e} x)$.
Therefore,$IF = e^{\log_{e}(\log_{e} x)} = \log_{e} x$.
The solution is $y \cdot IF = \int (Q \cdot IF) dx + C$.
$y \log_{e} x = \int \frac{1}{x} \log_{e} x dx$.
Let $v = \log_{e} x$,then $dv = \frac{1}{x} dx$. The integral becomes $\int v dv = \frac{v^2}{2} + C = \frac{(\log_{e} x)^2}{2} + C$.
So,$y \log_{e} x = \frac{(\log_{e} x)^2}{2} + C$.
Given $y = 1$ when $x = e$,we have $1 \cdot \log_{e} e = \frac{(\log_{e} e)^2}{2} + C$.
$1 = \frac{1}{2} + C \implies C = \frac{1}{2}$.
Substituting $C$ back,$y \log_{e} x = \frac{(\log_{e} x)^2}{2} + \frac{1}{2}$.
Dividing by $\log_{e} x$ (or multiplying by $2$),we get $2y \log_{e} x = (\log_{e} x)^2 + 1$,which simplifies to $2y = \log_{e} x + \frac{1}{\log_{e} x}$.