Question

Solve the given differential equation:
y' + 4y/x = x3y3​

Answer 1

Differential Equations

Solution:

The given differential equation is

$\quad \frac{dy}{dx}+4\frac{y}{x}=x^{3}y^{3}$

$\Rightarrow y^{-3}\frac{dy}{dx}+4\frac{y^{-2}}{x}=x^{3}$

To reduce it in linear form, we put $y^{-2}=z$ such that

$\quad -2y^{-3}\frac{dy}{dx}=\frac{dz}{dx}$

$\Rightarrow y^{-3}\frac{dy}{dx}=-\frac{1}{2}\frac{dz}{dx}$

So the given equation reduces to

$\quad -\frac{1}{2}\frac{dz}{dx}+4\frac{z}{x}=x^{3}$

$\Rightarrow \frac{dz}{dx}-8\frac{z}{x}=-2x^{3}\quad.....(1)$

It is a linear equation.

$\therefore\text{I.F.}=e^{-8\int \frac{dx}{x}}$

$\quad\quad\quad=e^{-8\:logx}$

$\quad\quad\quad=e^{log(x^{-8})}$

$\quad\quad\quad=x^{-8}$

Now multiplying $(1)$ by I.F., we get

$\quad\frac{d}{dx}(zx^{-8})=-2x^{-5}$

$\Rightarrow d(zx^{-8})=-2x^{-5}dx$

On integrated, we have

$\quad\int d(zx^{-8})=-2\int x^{-5}dx$

$\Rightarrow zx^{-8}=-2*\frac{x^{-4}}{-4}+C$

where $C$ is constant of integrated

$\Rightarrow \boxed{x^{-8}y^{-2}=\frac{1}{2}x^{-4}+C}$

This is the required integral.

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