How Do You Solve a First Order Differential Equation with an Initial Condition?

[DryRun]

Homework Statement
Solve $$\frac{dy(t)}{dt}+2ty(t)=5t<br /> \\y(0)=1$$

The attempt at a solution
Since this is a 1st order ODE, i find the integrating factor,
$$μ(t)=e^{\int P(t).dt}<br /> \\P(t)=2t<br /> \\Q(t)=5t<br /> \\μ(t)=e^{t^2}$$
Using formula:
$$\frac{dμy}{dt}=μQ$$
Integrating both R.H.S. and L.H.S.:
$$μy=\int e^{t^2}.5t\,.dt<br /> \\μy=\frac{5}{2}e^{t^2}+A<br /> \\y=\frac{5}{2}+\frac{A}{e^{t^2}}$$
Using y(0)=1
When t=0, y=1
$1=\frac{5}{2}+A$, so A=1-(5/2)=-3/2

Therefore, the particular solution is: y=(5/2)-(3/2) which gives y=1.

Is this answer correct?

 

[jgens]

Is this answer correct?

It looks correct to me.