Question

Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any.

Answer 1

Answer:

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$

Step-by-step explanation:

We are given that a differential equation

$(x^2+10)y'+xy-4x=0$

We have to find the general solution of given differential equation

$y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0$

$y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}$

Compare with

$y'+P(x) y=Q(x)$

We get

$P(x)=\frac{x}{x^2+10}$

$Q(x)=\frac{4x}{x^2+10}$

I.F=$e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}$

$e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C$

$y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C$

$y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C$

$y(x)=4+\frac{C}{\sqrt{x^2+10}}$