Question

Solve the linear differential equation 2xy' + y = 2√x

Answer 1

Answer:

Step-by-step explanation:

General form of the linear differential equation can be written as:

$\frac{dy}{dx}+P(x)y=Q(x)$

For this case, we can rewrite the equation as:

$\frac{dy}{dx}+\frac{1}{2x}y=\frac{\sqrt{x}}{x}$

Here $P(x) =\frac{1}{2x}; Q(x)=\frac{\sqrt{x}}{x}$

To find the solution (y(x)), we can use the integration factor method:

$Fy(x)=\int Q(x)Fdx+C \rightarrow F=e^{\int P(x)dx$

Then $F=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}\ln|x|\right}=\sqrt{|x|}$

So, we can find:

$y\sqrt{|x|}=\int \frac{\sqrt{x}\sqrt{|x|}}{x}dx+C$

Suppose that $x\in \double R$, then $\sqrt{|x|}=\sqrt{x}$ , and we find:

$y\sqrt{x}=x+C \rightarrow y(x)=\sqrt{x}+\frac{C\sqrt{x}}{x}$

To check our solution is right or not, put your y(x) back to the ODE:

$y' = \frac{1}{2\sqrt{x}}-\frac{C}{2\sqrt{x^{3}}}$

$2xy'=\frac{x-C}{\sqrt{x}}$

$2xy'+y=\frac{x-C}{\sqrt{x}}+\sqrt{x}+\frac{C\sqrt{x}}{x}=2\sqrt{x}$

(it means your solution is right)