Answer:
[tex]y(x)=4+\frac{C}{\sqrt{x^2+10}}[/tex]
Step-by-step explanation:
We are given that a differential equation
[tex](x^2+10)y'+xy-4x=0[/tex]
We have to find the general solution of given differential equation
[tex]y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0[/tex]
[tex]y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}[/tex]
Compare with
[tex]y'+P(x) y=Q(x)[/tex]
We get
[tex]P(x)=\frac{x}{x^2+10}[/tex]
[tex]Q(x)=\frac{4x}{x^2+10}[/tex]
I.F=[tex]e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}[/tex]
[tex]e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}[/tex]
[tex]y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C[/tex]
[tex]y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C[/tex]
[tex]y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C[/tex]
[tex]y(x)=4+\frac{C}{\sqrt{x^2+10}}[/tex]
