Question

The solution of the differential equation $x^3\frac{dy}{dx}+4x^2\tan y=e^x\sec y$ satisfying $y\left(1\right)=0$ is

• A. $\tan y=(x-2)e^x\log x$
• B. $\sin y=e^x(x-1)x^{-4}$
• C. $\tan y=(x-1)e^x{x}^{-3}$
• D. $\sin y=e^x(x-1)x^{-3}$

Answer 1

The correct option is B $\sin y=e^x(x-1)x^{-4}$

Given,

$x^3\frac{dy}{dx}+4x^{2}tany=e^{x}secy$

Dividing whole equation by $x^{3}secy$

$cosy\frac{dy}{dx}+\frac{4}{x}siny=\frac{e^x}{x^3}$

let siny=t

$cosydy=dt$

Put in above equation

$\frac{dt}{dx}+\frac{4}{x}t=\frac{e^x}{x^3}$

P=$\frac{4}{x},Q=\frac{e^x}{x^3}$

I.f.=$e^{\int P.dx}$

=$e^{\int \frac{4}{x}dx}$

=$e^{4logx}$

=$e^{log{x}^4}$

=$x^4$

Complete solution,

$t\times x^4=\int x^4\times \frac{e^x}{x^3}dx+c$

$t\times x^4=\int x{e}^{x}dx+c$

$t\times x^4=x{e}^x-e^x+c$

$siny\times x^4=x{e}^x-e^x+c$

at x=1,y=0

$0=1.e-e+c$

c=0

$siny\times x^4=e^x(x-1)+0$

$siny=x^{-4}{e}^x(x-1)$