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

Question

asked Jun 18, 2023 45.9k views

1 Answer

ITrout · Answer 1 · 2023-06-22T04:00:59+0000

Answer:

Explanation:

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

(dy)/(dx)+P(x)y=Q(x)

For this case, we can rewrite the equation as:

(dy)/(dx)+(1)/(2x)y=(√(x))/(x)

Here
P(x) =(1)/(2x); Q(x)=(√(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
$F=e^{\int (1)/(2x)dx}=e^x=√(|x|)$

So, we can find:

$y√(|x|)=\int (√(x)√(|x|))/(x)dx+C$

Suppose that
$x\in \double R$ , then
√(|x|)=√(x) , and we find:

$y√(x)=x+C \rightarrow y(x)=√(x)+(C√(x))/(x)$

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

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

2xy'=(x-C)/(√(x))

2xy'+y=(x-C)/(√(x))+√(x)+(C√(x))/(x)=2√(x)

(it means your solution is right)