Question

Chapter: Differential Equations Please solve using all the necessary steps.

Answer 1

Given equation :-

$\displaystyle \sf{x=1+xy\frac{dy}{dx} +\frac{x^2y^2}{2!}\bigg(\frac{dy}{dx} \bigg)^2 +\frac{x^3y^3}{3!}\bigg(\frac{dy}{dx} \bigg)^3...}$

To Find :-

The solution of the given differential equation.

Solution :-

We know from maclaurin series expansion for powers of e :-

$\boxed{\bf{e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+... }}\:\:\:\:\cdots \sf{(i)}$

Comparing equation (i) with the given equation, we get :

$\sf{x=e^{xy\frac{dy}{dx} }}$

$\sf{\longrightarrow log_ex=xy\dfrac{dy}{dx} }$

$\sf{\longrightarrow ydy=\dfrac{log_e x}{x} dx}$

Integrating both the sides :-

$\displaystyle\sf{\longrightarrow \int ydy=\int\frac{log_ex}{x}dx }$

$\sf{\longrightarrow \dfrac{y^2}{2}= \dfrac{(log_ex)^2}{2}+C }$

$\sf{\longrightarrow y^2=(log_ex)^2+C}$

$\sf{\longrightarrow y=\pm\sqrt{(log_ex)^2}+C }$

$\sf{\longrightarrow y=\pm log_ex^2+C}$

◘ For provided options, the required answer is :-

B. y² = (log x)² + c

Hope this helps :))))

Answer 2

Given,

$\displaystyle\longrightarrow x=1+xy\cdot\dfrac{dy}{dx}+\dfrac{x^2y^2}{2!}\left(\dfrac{dy}{dx}\right)^2+\dfrac{x^3y^3}{3!}\left(\dfrac{dy}{dx}\right)^3+\,\dots$

Or,

$\displaystyle\longrightarrow x=\sum_{k=0}^{\infty}\dfrac{x^ky^k}{k!}\left(\dfrac{dy}{dx}\right)^k$

$\displaystyle\longrightarrow x=\sum_{k=0}^{\infty}\dfrac{\left(xy\cdot\dfrac{dy}{dx}\right)^k}{k!}\quad\quad\dots(1)$

But we know that,

$\displaystyle\longrightarrow e^x=\sum_{k=0}^{\infty}\dfrac{x^k}{k!}$

Thus (1) becomes,

$\displaystyle\longrightarrow x=e^{xy\cdot\frac{dy}{dx}}$

Taking log,

$\displaystyle\longrightarrow\log x=xy\cdot\dfrac{dy}{dx}$

$\displaystyle\longrightarrow y\ dy=\dfrac{1}{x}\,\log x\ dx$

Now integrating,

$\displaystyle\longrightarrow\int y\ dy=\int\dfrac{1}{x}\,\log x\ dx$

Integrating RHS by parts,

$\displaystyle\longrightarrow\dfrac{y^2}{2}=\log x\cdot\log x-\int\dfrac{1}{x}\,\log x\ dx+C$

$\displaystyle\longrightarrow\dfrac{y^2}{2}=(\log x)^2-\int y\ dy+C$

$\displaystyle\longrightarrow\dfrac{y^2}{2}=(\log x)^2-\dfrac{y^2}{2}+C$

$\displaystyle\longrightarrow\underline{\underline{y^2=(\log x)^2+C}}$

Hence (B) is the answer.