Answer:
Step-by-step explanation:
General form of the linear differential equation can be written as:
[tex]\frac{dy}{dx}+P(x)y=Q(x)[/tex]
For this case, we can rewrite the equation as:
[tex]\frac{dy}{dx}+\frac{1}{2x}y=\frac{\sqrt{x}}{x}[/tex]
Here [tex]P(x) =\frac{1}{2x}; Q(x)=\frac{\sqrt{x}}{x}[/tex]
To find the solution (y(x)), we can use the integration factor method:
[tex]Fy(x)=\int Q(x)Fdx+C \rightarrow F=e^{\int P(x)dx[/tex]
Then [tex]F=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}\ln|x|\right}=\sqrt{|x|}[/tex]
So, we can find:
[tex]y\sqrt{|x|}=\int \frac{\sqrt{x}\sqrt{|x|}}{x}dx+C[/tex]
Suppose that [tex]x\in \double R[/tex], then [tex]\sqrt{|x|}=\sqrt{x}[/tex] , and we find:
[tex]y\sqrt{x}=x+C \rightarrow y(x)=\sqrt{x}+\frac{C\sqrt{x}}{x}[/tex]
To check our solution is right or not, put your y(x) back to the ODE:
[tex]y' = \frac{1}{2\sqrt{x}}-\frac{C}{2\sqrt{x^{3}}}[/tex]
[tex]2xy'=\frac{x-C}{\sqrt{x}}[/tex]
[tex]2xy'+y=\frac{x-C}{\sqrt{x}}+\sqrt{x}+\frac{C\sqrt{x}}{x}=2\sqrt{x}[/tex]
(it means your solution is right)
