Variable separable type differential equations Questions in English

(B) Given the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$.
Let $v = 2x-3y$. Then $\frac{dv}{dx} = 2 - 3\frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{3}(2 - \frac{dv}{dx})$.
Substituting into the equation: $\frac{1}{3}(2 - \frac{dv}{dx}) = \frac{v+5}{3v+7}$.
$2 - \frac{dv}{dx} = \frac{3v+15}{3v+7} \implies \frac{dv}{dx} = 2 - \frac{3v+15}{3v+7} = \frac{6v+14-3v-15}{3v+7} = \frac{3v-1}{3v+7}$.
Separating variables: $\int \frac{3v+7}{3v-1} dv = \int dx$.
$\int (1 + \frac{8}{3v-1}) dv = \int dx \implies v + \frac{8}{3} \log |3v-1| = x + c$.
Substituting $v = 2x-3y$: $(2x-3y) + \frac{8}{3} \log |3(2x-3y)-1| = x + c$.
$x - 3y + \frac{8}{3} \log |6x-9y-1| + c = 0$.

(D) Given the differential equation: $\cos y \frac{dy}{dx} = e^x e^{\sin y} + x^2 e^{\sin y}$.
Divide both sides by $e^{\sin y}$: $e^{-\sin y} \cos y \frac{dy}{dx} = e^x + x^2$.
Let $u = \sin y$,then $\frac{du}{dx} = \cos y \frac{dy}{dx}$.
The equation becomes $e^{-u} \frac{du}{dx} = e^x + x^2$.
Integrating both sides with respect to $x$: $\int e^{-u} du = \int (e^x + x^2) dx$.
$-e^{-u} = e^x + \frac{x^3}{3} + C_1$.
Substitute $u = \sin y$ back: $-e^{-\sin y} = e^x + \frac{x^3}{3} + C_1$.
Rearranging to the form $f(x) + e^{-\sin y} = C$: $e^x + \frac{x^3}{3} + e^{-\sin y} = C$.
Comparing this with $f(x) + e^{-\sin y} = C$,we get $f(x) = e^x + \frac{x^3}{3}$.

(A) Given the differential equation: $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$
We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.
Substituting this into the equation,we get: $\frac{d(xy)}{dx} = \frac{x f(xy)}{f'(xy)}$.
Rearranging the terms to separate the variables $xy$ and $x$: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.
Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.
This yields: $\ln |f(xy)| = \frac{x^2}{2} + C$.
Taking the exponential of both sides: $|f(xy)| = e^{\frac{x^2}{2} + C} = e^C \cdot e^{x^2 / 2}$.
Letting $k = e^C$,we get: $|f(xy)| = k e^{x^2 / 2}$.

(D) Given equation: $(x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0$
Let $x+y=t$. Then $1+\frac{d y}{d x}=\frac{d t}{d x}$,which implies $\frac{d y}{d x}=\frac{d t}{d x}-1$.
Substituting this into the given equation:
$t^{2}(\frac{d t}{d x}-1)=a^{2}$
$t^{2} \frac{d t}{d x} = a^{2}+t^{2}$
Separating the variables:
$\frac{t^{2}}{a^{2}+t^{2}} d t = d x$
Integrating both sides:
$\int \frac{t^{2}+a^{2}-a^{2}}{t^{2}+a^{2}} d t = \int d x$
$\int (1 - \frac{a^{2}}{t^{2}+a^{2}}) d t = x + C'$
$t - a^{2} \cdot \frac{1}{a} \tan^{-1}(\frac{t}{a}) = x + C'$
$t - a \tan^{-1}(\frac{t}{a}) = x + C'$
Substituting $t = x+y$ back:
$(x+y) - a \tan^{-1}(\frac{x+y}{a}) = x + C'$
$y - C' = a \tan^{-1}(\frac{x+y}{a})$
$\frac{y-C'}{a} = \tan^{-1}(\frac{x+y}{a})$
$\tan(\frac{y-C'}{a}) = \frac{x+y}{a}$
Let $C = -C'$. Then the general solution is $\tan(\frac{y+C}{a}) = \frac{x+y}{a}$.

(A) Given the differential equation: $(x+y)^{2} \frac{dy}{dx} = a^{2}$.
Let $v = x+y$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting these into the original equation: $v^{2} (\frac{dv}{dx} - 1) = a^{2}$.
Rearranging the terms: $v^{2} \frac{dv}{dx} = v^{2} + a^{2}$,so $\frac{dv}{dx} = \frac{v^{2} + a^{2}}{v^{2}}$.
Separating the variables: $\frac{v^{2}}{v^{2} + a^{2}} dv = dx$.
Integrating both sides: $\int \frac{v^{2}}{v^{2} + a^{2}} dv = \int dx$.
This can be written as: $\int (1 - \frac{a^{2}}{v^{2} + a^{2}}) dv = x + C'$.
Integrating gives: $v - a \tan^{-1}(\frac{v}{a}) = x + C'$.
Substituting $v = x+y$: $(x+y) - a \tan^{-1}(\frac{x+y}{a}) = x + C'$.
Simplifying: $y - a \tan^{-1}(\frac{x+y}{a}) = C'$.
Rearranging: $\frac{y-C'}{a} = \tan^{-1}(\frac{x+y}{a})$.
Taking the tangent on both sides: $\tan(\frac{y-C'}{a}) = \frac{x+y}{a}$.
Letting $-C' = C$,we get $\frac{x+y}{a} = \tan(\frac{y+C}{a})$.

(A) Given the differential equation: $\log_{e}\left(\frac{dy}{dx}\right) = x + y$.
By the definition of logarithm,we can write this as: $\frac{dy}{dx} = e^{x+y}$.
Using the property of exponents,we have: $\frac{dy}{dx} = e^x \cdot e^y$.
Separating the variables,we get: $e^{-y} dy = e^x dx$.
Integrating both sides: $\int e^{-y} dy = \int e^x dx$.
This yields: $-e^{-y} = e^x + C_1$.
Rearranging the terms,we get: $e^x + e^{-y} = -C_1$.
Letting $-C_1 = C$,the general solution is: $e^x + e^{-y} = C$.

(D) Given differential equation is $x \, dy - y \, dx = 0$.
Rearranging the terms,we get $x \, dy = y \, dx$.
Separating the variables,we have $\frac{dy}{y} = \frac{dx}{x}$.
Integrating both sides,we get $\int \frac{dy}{y} = \int \frac{dx}{x} + C$.
This results in $\ln|y| = \ln|x| + \ln|c|$,where $\ln|c|$ is the constant of integration.
Using logarithmic properties,$\ln|y| = \ln|cx|$,which implies $y = cx$.
The equation $y = cx$ represents a straight line passing through the origin.

(B) Given the differential equation: $\frac{dy}{dx} = e^{y+x} + e^{y-x}$.
We can rewrite the right side as: $\frac{dy}{dx} = e^y(e^x + e^{-x})$.
Separating the variables,we get: $e^{-y} dy = (e^x + e^{-x}) dx$.
Integrating both sides: $\int e^{-y} dy = \int (e^x + e^{-x}) dx$.
This yields: $-e^{-y} = e^x - e^{-x} + c$.
Multiplying by $-1$,we obtain: $e^{-y} = e^{-x} - e^x + c$.
Thus,the correct option is $B$.

(D) Given differential equation: $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$
Let $x+y=v$. Then $1+\frac{d y}{d x}=\frac{d v}{d x}$,so $\frac{d y}{d x}=\frac{d v}{d x}-1$.
Substituting this into the equation: $\frac{d v}{d x}-1=\frac{v+1}{2 v+1}$
$\frac{d v}{d x}=\frac{v+1}{2 v+1}+1 = \frac{v+1+2 v+1}{2 v+1} = \frac{3 v+2}{2 v+1}$
Separating variables: $\frac{2 v+1}{3 v+2} d v=d x$
Rewrite the numerator: $\frac{\frac{2}{3}(3 v+2)-\frac{1}{3}}{3 v+2} d v=d x$
$\left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v=d x$
Integrating both sides: $\int \left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v = \int d x + C'$
$\frac{2}{3} v - \frac{1}{9} \log |3 v+2| = x + C'$
Substitute $v=x+y$: $\frac{2}{3}(x+y) - \frac{1}{9} \log |3 x+3 y+2| = x + C'$
Multiply by $9$: $6(x+y) - \log |3 x+3 y+2| = 9 x + 9 C'$
$6 x + 6 y - 9 x - \log |3 x+3 y+2| = C$
$-3 x + 6 y - \log |3 x+3 y+2| = C$
Multiplying by $-1$: $3 x - 6 y + \log |3 x+3 y+2| = C$
Thus,the correct option is $D$.

(A) Given the differential equation: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.
We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.
Substituting this into the equation,we get: $\frac{d(xy)}{dx} = x \frac{f(xy)}{f'(xy)}$.
Rearranging the terms to separate the variables,we have: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.
Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.
This yields: $\ln |f(xy)| = \frac{x^2}{2} + k$,where $k$ is the constant of integration.
Taking the exponential of both sides: $|f(xy)| = e^{\frac{x^2}{2} + k} = e^k \cdot e^{\frac{x^2}{2}}$.
Letting $C = e^k$,we get: $|f(xy)| = Ce^{\frac{x^2}{2}}$.

(A) Given the differential equation: $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \, dy = (1+2 \sin y) \, dx$.
Separating the variables,we get: $\int \frac{\cos y}{1+2 \sin y} \, dy = \int \frac{dx}{16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})}$.
Let $u = 1+2 \sin y$,then $du = 2 \cos y \, dy$,so $\int \frac{\cos y}{1+2 \sin y} \, dy = \frac{1}{2} \ln |1+2 \sin y|$.
For the $RHS$,let $t = 4+\sqrt{9+\sqrt{x}}$. Then $t-4 = \sqrt{9+\sqrt{x}}$. Squaring both sides: $(t-4)^2 = 9+\sqrt{x}$,so $\sqrt{x} = (t-4)^2 - 9$.
Differentiating $t = 4+\sqrt{9+\sqrt{x}}$,we get $dt = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{dx}{4\sqrt{x(9+\sqrt{x})}} = \frac{dx}{4\sqrt{x+9\sqrt{x}}}$.
Thus,$\frac{dx}{\sqrt{x+9\sqrt{x}}} = 4 \, dt$.
The integral becomes: $\frac{1}{2} \ln |1+2 \sin y| = \int \frac{4 \, dt}{16t} = \frac{1}{4} \ln |t| + C = \frac{1}{4} \ln |4+\sqrt{9+\sqrt{x}}| + C$.
Using $y(256) = \frac{\pi}{2}$: $\frac{1}{2} \ln(1+2 \sin \frac{\pi}{2}) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{256}}) + C \implies \frac{1}{2} \ln 3 = \frac{1}{4} \ln(4+\sqrt{9+16}) + C = \frac{1}{4} \ln 9 + C = \frac{1}{2} \ln 3 + C$. Thus $C = 0$.
Now,for $y(49) = \alpha$: $\frac{1}{2} \ln(1+2 \sin \alpha) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{49}}) = \frac{1}{4} \ln(4+\sqrt{16}) = \frac{1}{4} \ln 8 = \frac{1}{4} \ln(2^3) = \frac{3}{4} \ln 2$.
So $\ln(1+2 \sin \alpha) = \frac{3}{2} \ln 2 = \ln(2^{3/2}) = \ln(2\sqrt{2})$.
Therefore,$1+2 \sin \alpha = 2\sqrt{2}$,which gives $2 \sin \alpha = 2\sqrt{2}-1$.

(A) Given the differential equation: $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$.
Divide both sides by $x^{2}$ (since $x > 0$): $\frac{x dy - y dx}{x^{2}} = \frac{\sqrt{x^{2} + y^{2}}}{x^{2}} dx$.
This simplifies to: $d\left(\frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^{2}} \cdot \frac{1}{x} dx$.
Integrating both sides: $\int \frac{d(\frac{y}{x})}{\sqrt{1 + (\frac{y}{x})^{2}}} = \int \frac{1}{x} dx$.
Using the standard integral $\int \frac{du}{\sqrt{1 + u^{2}}} = \ln|u + \sqrt{1 + u^{2}}| + C$,we get: $\ln\left(\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}}\right) = \ln x + C$.
Given $y(1) = 0$,substitute $x = 1, y = 0$: $\ln(0 + \sqrt{1 + 0}) = \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus,$\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} = x$.
Multiplying by $x$: $y + \sqrt{x^{2} + y^{2}} = x^{2}$.
To find $y(3)$,substitute $x = 3$: $y + \sqrt{9 + y^{2}} = 9$.
$\sqrt{9 + y^{2}} = 9 - y$.
Squaring both sides: $9 + y^{2} = 81 - 18y + y^{2}$.
$18y = 72 \Rightarrow y = 4$.

(B) Given the differential equation $\frac{dy}{dx} = e^{x-y}$.
Using the property of exponents,we can write $\frac{dy}{dx} = \frac{e^x}{e^y}$.
By separating the variables,we get $e^y \, dy = e^x \, dx$.
Integrating both sides,we have $\int e^y \, dy = \int e^x \, dx$.
This results in $e^y = e^x + C$,where $C$ is the constant of integration.
Rearranging the terms,we get $e^x - e^y = -C$,which can be written as $e^x - e^y = c$ (where $c = -C$ is an arbitrary constant).

(B) The given differential equation is $\frac{dy}{dx} = e^{x+y} = e^x \cdot e^y$.
By separating the variables,we get $e^{-y} \, dy = e^x \, dx$.
Integrating both sides,we have $\int e^{-y} \, dy = \int e^x \, dx$.
This results in $-e^{-y} = e^x + C'$,where $C'$ is the constant of integration.
Rearranging the terms,we get $e^x + e^{-y} = -C'$.
Letting $C = -C'$,the general solution is $e^x + e^{-y} = C$.

(D) The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.
Rearranging the terms to separate the variables,we get $\frac{dy}{y+1} = -\frac{\cos x}{1+\sin x} dx$.
Integrating both sides,we have $\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx$.
This yields $\ln|y+1| = -\ln|1+\sin x| + C$,which simplifies to $\ln|y+1| + \ln|1+\sin x| = C$.
Using the property of logarithms,we get $(y+1)(1+\sin x) = K$,where $K = e^C$.
Given the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$ into the equation: $(0+1)(1+\sin 0) = K$,which gives $1(1+0) = K$,so $K = 1$.
Thus,the particular solution is $(y+1)(1+\sin x) = 1$.
If the curve passes through $(\alpha, -\frac{1}{2})$,we substitute $x = \alpha$ and $y = -\frac{1}{2}$ into the equation:
$(-\frac{1}{2} + 1)(1+\sin \alpha) = 1$.
$\frac{1}{2}(1+\sin \alpha) = 1$.
$1+\sin \alpha = 2$.
$\sin \alpha = 1$.
Therefore,$\alpha = \frac{\pi}{2}$.

(C) Given the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$.
Separating the variables,we get $\int \frac{dy}{1-y^2} = \int (1+x^2) dx$.
Integrating both sides,we have $\frac{1}{2} \ln|\frac{1+y}{1-y}| = x + \frac{x^3}{3} + C$.
Using the initial condition $y(0) = \frac{1}{2}$,we find $C$: $\frac{1}{2} \ln|\frac{1+1/2}{1-1/2}| = 0 + 0 + C \Rightarrow C = \frac{1}{2} \ln(3)$.
Substituting $C$ back,$\frac{1}{2} \ln|\frac{1+y}{1-y}| = x + \frac{x^3}{3} + \frac{1}{2} \ln(3) \Rightarrow \ln|\frac{1+y}{1-y}| = 2(x + \frac{x^3}{3}) + \ln(3)$.
At $x=1$,$\ln|\frac{1+y(1)}{1-y(1)}| = 2(1 + \frac{1}{3}) + \ln(3) = \frac{8}{3} + \ln(3)$.
Thus,$\frac{1+y(1)}{1-y(1)} = 3e^{8/3}$.
Let $k = 3e^{8/3}$. Then $1+y(1) = k - ky(1) \Rightarrow y(1)(1+k) = k-1 \Rightarrow y(1) = \frac{k-1}{k+1}$.
$2y(1)-1 = 2(\frac{k-1}{k+1}) - 1 = \frac{2k-2-k-1}{k+1} = \frac{k-3}{k+1}$.
Substituting $k = 3e^{8/3}$,we get $\frac{3e^{8/3}-3}{3e^{8/3}+1}$. This expression simplifies to the form involving $\tan$ based on the hyperbolic identity $\tanh(u) = \frac{e^{2u}-1}{e^{2u}+1}$.

(C) Given the differential equation: $\frac{dy}{y} = (2 + \ln x) dx$.
Integrating both sides: $\int \frac{dy}{y} = \int (2 + \ln x) dx$.
$\ln y = 2x + (x \ln x - x) + C = x \ln x + x + C$.
Since the curve passes through $(1, e)$,substitute $x = 1$ and $y = e$: $\ln e = 1 \ln 1 + 1 + C$.
$1 = 0 + 1 + C$,which gives $C = 0$.
Thus,the equation of the curve is $\ln y = x \ln x + x$.
To find $f(e)$,substitute $x = e$: $\ln f(e) = e \ln e + e = e(1) + e = 2e$.
Therefore,$f(e) = e^{2e}$.

(C) The given differential equation is $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$.
Separating the variables,we get $\int \frac{dy}{y^2-y+1} = \int (1+x^2) dx$.
Completing the square in the denominator: $y^2-y+1 = (y-1/2)^2 + 3/4$.
Thus,$\int \frac{dy}{(y-1/2)^2 + (\sqrt{3}/2)^2} = x + \frac{x^3}{3} + C$.
Using the formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a})$,we get $\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y-1}{\sqrt{3}}\right) = x + \frac{x^3}{3} + C$.
Given $y(0) = 1/2$,we have $\frac{2}{\sqrt{3}} \tan^{-1}(0) = 0 + 0 + C$,which implies $C = 0$.
So,$\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y-1}{\sqrt{3}}\right) = x + \frac{x^3}{3}$.
At $x = 1$,$\frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2y(1)-1}{\sqrt{3}}\right) = 1 + 1/3 = 4/3$.
$\tan^{-1} \left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{4}{3} \cdot \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.
Therefore,$\frac{2y(1)-1}{\sqrt{3}} = \tan \left(\frac{2}{\sqrt{3}}\right)$,which gives $2y(1)-1 = \sqrt{3} \tan \left(\frac{2}{\sqrt{3}}\right)$.
Note: The provided options appear to have a calculation discrepancy in the argument of the tangent function. Based on the standard derivation,the correct value is $\sqrt{3} \tan \left(\frac{2}{\sqrt{3}}\right)$.

(C) Given the differential equation: $\frac{dy}{y} = (2 + \log_e x) dx$.
Integrating both sides: $\int \frac{dy}{y} = \int (2 + \log_e x) dx$.
$\log_e y = 2x + (x \log_e x - x) + C = x \log_e x + x + C$.
Since the curve passes through $(1, e)$,substitute $x = 1$ and $y = e$:
$\log_e e = 1 \cdot \log_e 1 + 1 + C$.
$1 = 0 + 1 + C \Rightarrow C = 0$.
Thus,the equation of the curve is $\log_e y = x \log_e x + x$.
To find $y(e)$,substitute $x = e$:
$\log_e y = e \log_e e + e = e(1) + e = 2e$.
Therefore,$y = e^{2e}$.