Question

Solve : $\frac{dy}{dx}=(y-4x{)}^2$

• A. $\frac{y-4x-2}{y-4x+2}=c{e}^{-4x}$
• B. $\frac{y-4x+2}{y-4x-2}=c{e}^{4x}$
• C. $\frac{y-4x+2}{y-4x-2}=c{e}^{-4x}$
• D. $\frac{y-4x-2}{y-4x+2}=c{e}^{4x}$

Answer 1

The correct option is C $\frac{y-4x+2}{y-4x-2}=c{e}^{-4x}$

Given, $\frac{dy}{dx}={(y-4x)}^2$

Substituting $v=y-4x\Rightarrow \frac{dv}{dx}=\frac{dy}{dx}-4$

$\frac{dv}{dx}+4=v^2\Rightarrow \frac{dv}{dx}=v^2-4\Rightarrow \frac{\frac{dv}{dx}}{v^2-4}=1$

Integrating both sides w.r.t x, we get

$\int \frac{\frac{dv}{dx}}{v^2-4}dx=\int dx\int \frac{1}{v^2-4}dv=\int dx$

We have,

$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+c$

$\Rightarrow \frac{1}{4}\log \left|\frac{v-2}{v+2}\right|+c=x+c\Rightarrow \frac{y-4x+2}{y-4x-2}=c{e}^{-4x}$