Question

plzzzzz help me... ​

Answer 1

$\large\underline{\purple{\sf \orange{\bigstar}Given.equation ::-}}$

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

$\large\underline{\purple{\sf \red{\bigstar}To.Find::-}}$

The solution of the given differential equation.

$\large\underline{\pink{\sf \green{\bigstar}Solution ::-}}$

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

Answer 2

$\large\bold{\underline{\underline{Given \: Equation:-}}}$

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

$\large\underline{\pink{\sf \purple{\bigstar}To.Find:-}}$

The solution of the given differential equation.

$\large\underline{\green{\sf \red{\bigstar}Solution :-}}$

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