I am given these two things: \begin{cases} y'=xy \\ (x_0,y_0)=(0,1) \end{cases}

The first part of the problem is to show how to verify that the general solution is $y=ce^{x^2/2}$, which I did (I'm putting this here because I'm not sure if its necessary to know this for my question).

Now I need to find the solution to the IVP at $x=0.1$ and $x=0.2$, but I'm not sure how to go about doing that. From the problems I've had to solve before I am given a specific condition, such as $y(0)=1$, but I don't have that here. What do I need to do?

Edit: Looking again at the $(x_0,y_0)=(0,1)$ I see that I am given the initial condition, which is $y(0)=1$. Is that correct?


1 Answer 1


Yes $(x_0,y_0)=(0,1)$ is the same as $y(0)=1$. Thus using this for your general solution $f(x) = c e^{\frac12 x^2}$ we hace that $c=1$. Thus the solution to the IVP is $f(x)=e^{\frac12 x^2}$. Now we evaluate this solution af $x=0.1$ and get $f(0.1) = e^{0.005}$. Next we evaluate at $x=0.2$ getting $f(0.2)= e^{0.02}$.


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