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Solution of differential equation x^{2}=1+\left ( \frac{x}{y} \right )^{-1}\frac{dy}{dx}+\frac{\left ( \frac{x}{y} \right )^{-2}\left ( \frac{dy}{dx} \right )^{2}}{2!}+\frac{\left ( \frac{x}{y} \right )^{-3}\left ( \frac{dy}{dx} \right )^{3}}{3!}+....  is 

Option: 1

y^{2}=x^{2}(lnx^{2}-1)+c


Option: 2

y=x^{2}(lnx-1)+c


Option: 3

y^{2}=x(lnx-1)+c


Option: 4

y=x^{2}+c


Answers (1)

best_answer

 

Solution of Differential Equation -

\frac{\mathrm{d}y }{\mathrm{d} x} =f\left ( ax+by+c \right )

put

 Z =ax+by+c

 

 

- wherein

Equation with convert to

\int \frac{dz}{bf\left ( z \right )+a} =x+c

 

 

 

 

x^{2}=e^{\left ( \frac{x}{y} \right )^{-1}\left ( \frac{dy}{dx} \right )}\; \; \; \; \Rightarrow \; \; \; x^{2}=e^{\left ( \frac{x}{y} \right )\left ( \frac{dy}{dx} \right )}

\Rightarrow In\; x^{2}=\frac{y}{x}\; \frac{dy}{dx}

\Rightarrow\: \int xlnx^{2}dx=\int y\; dy

Put\: x^{2}=t\; \; \; \; \Rightarrow 2xdx=dt

\frac{1}{2}\int lnt\: dt=\frac{y^{2}}{2}

\Rightarrow c+tln\: t-t=y^{2}

\Rightarrow y^{2}=x^{2}\: lnx^{2}-x^{2}+c

 

Posted by

Pankaj Sanodiya

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