Question

A solution of differential equation is a relation between the variables (independent and dependent), which is free of derivatives of any order and satisfies the differential equation identically.
$$\begin{array}{cc}\text{Differential equation}& \text{General solution}\\ \text{(A)}2x{y}^{2}dx=e^x(dy-ydx)& \text{(P)}4x^{3}y+6e^x-3y^3=cy\\ \text{(B)}y(2x^{2}y+e^x)dx=(e^x+y^3)dy& \text{(Q)}x+{\log }_e(x^2+y^2)=c\\ \text{(C)}(x^2+y^2+2x)dx+2ydy=0& \text{(R)}{x}^2{\log }_e\left(x^2{y}^2\right)+y^3=c{x}^2\\ \text{(D)}(2x^{2}y-2y^4)dx+(2x^3+3x{y}^3)dy=0& \text{(S)}{x}^{2}y+e^x=cy\end{array}$$
Here, $c$ is an integration constant.

Which of the following is the INCORRECT combination?

• A. $\left(D\right)\to \left(Q\right)$
• B. $\left(B\right)\to \left(P\right)$
• C. $\left(A\right)\to \left(S\right)$
• D. $\left(C\right)\to \left(Q\right)$

Answer 1

The correct option is A $\left(D\right)\to \left(Q\right)$

$\left(A\right)$

$2x{y}^{2}dx=e^x(dy-ydx)\Rightarrow \frac{2x}{e^x}{y}^2=\frac{dy}{dx}-y\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=2x{e}^{-x}$
Taking $\frac{-1}{y}=t\Rightarrow \frac{dt}{dx}=\frac{1}{y^2}\frac{dy}{dx}$
$\Rightarrow \frac{dt}{dx}+t=2x{e}^{-x}$

$I.F.=e^{\int dx}=e^x$
$\Rightarrow t{e}^x=\int 2xdx$
$\Rightarrow -\frac{e^x}{y}=x^2+c_1$
$\Rightarrow x^{2}y+e^x=cy$, where $c=-c_1$
$\left(A\right)\to \left(S\right)$

$\left(B\right)$

$y(2x^{2}y+e^x)dx=(e^x+y^3)dy\Rightarrow 2x^2{y}^{2}dx-y^{3}dy+y{e}^{x}dx-e^{x}dy=0\Rightarrow (2x^{2}dx-ydy)+\frac{e^x(ydx-dy)}{y^2}=0$
$\Rightarrow (2x^{2}dx-ydy)+d\left(\frac{e^x}{y}\right)=0$
On integration, we get
$\Rightarrow \frac{2x^3}{3}-\frac{y^2}{2}+\frac{e^x}{y}=c_2$
$\Rightarrow 4x^{3}y+6e^x-3y^3=cy$, where $c=6c_2$
$\left(B\right)\to \left(P\right)$

$\left(C\right)$

$(x^2+y^2+2x)dx+2ydy=0\Rightarrow x^2+y^2+2x+2y\frac{dy}{dx}=0\Rightarrow (x^2+y^2)+\frac{d}{dx}(x^2+y^2)=0\Rightarrow \frac{d(x^2+y^2)}{x^2+y^2}=-dx$
On integration, we get
${\log }_e(x^2+y^2)=-x+c\Rightarrow x+{\log }_e(x^2+y^2)=c$
$\left(C\right)\to \left(Q\right)$

$\left(D\right)$

$(2x^{2}y-2y^4)dx+(2x^3+3x{y}^3)dy=0\Rightarrow 2x^{2}y-2y^4+2x^3\frac{dy}{dx}+3x{y}^3\frac{dy}{dx}=0\Rightarrow 2x^2(y+x\frac{dy}{dx})+y^3(3x\frac{dy}{dx}-2y)=0\Rightarrow 2(y+x\frac{dy}{dx})+y(\frac{3y^2}{x}\frac{dy}{dx}-\frac{2y^3}{x^2})=0\Rightarrow 2(1+\frac{x}{y}\frac{dy}{dx})+(\frac{3y^2}{x}\frac{dy}{dx}-\frac{2y^3}{x^2})=0\Rightarrow \frac{2}{x}+\frac{2}{y}\frac{dy}{dx}+(\frac{3y^2}{x^2}\frac{dy}{dx}-\frac{2y^3}{x^3})=0\Rightarrow \frac{2}{x}+\frac{2}{y}\frac{dy}{dx}+\frac{d}{dx}\left(\frac{y^3}{x^2}\right)=0\Rightarrow 2\frac{dx}{x}+2\frac{dy}{y}+d\left(\frac{y^3}{x^2}\right)=0$
On integration, we get
$\ln x^2+\ln y^2+\frac{y^3}{x^2}=c\Rightarrow x^2{\log }_e\left(x^2{y}^2\right)+y^3=c{x}^2$
$\left(D\right)\to \left(R\right)$