Solving the following Initial Value Problem: $y'=\frac{x}{1+y^2}$, $y(-1)=1$.

Question

I'm new to this and I'm stuck on one of my problems in the book. $y'=\frac{x}{1+y^2}$ with $y(-1)=1$.

I know I need to find $P(x)$ and then integrate $e^{\int P(x)}$. I am thinking that my $P(x) = x$ here, so when I integrate I have $e^\frac{x^2}{2}$.

However, I'm lost on what to do after. Can someone guide me through this example and show what to do with the $y(-1)=1$?

Answer 1

This is a standard exercise in what's called separable ODE. Write $\dfrac{dy}{dx} = \dfrac{x}{1+y^2}\implies \int(1+y^2)dy = \int xdx\implies y+\frac{y^3}{3} = x^2+C.$ This is the general formula for your function and it is an implicit one, so you cannot hope to find a better looking formula unless you want to solve a cubic equation. Now, simply use the initial condition to find what $C$ is.

Answer 2

Another hint

$$y'=\frac{x}{1+y^2}$$ $$1+y^2=xx'$$ $$\frac 12 (x^2)'=1+y^2$$ $$\frac 12 x^2=\int 1+y^2dy= \frac {y^3} 3 +y+K$$ $$\frac {x^2}2= \frac {y^3} 3 +y+K$$ $$y(-1)=1 \implies \frac {1}2= \frac {1} 3 +1+K \implies K=- \frac 56$$ $$\boxed {\frac {x^2}2= \frac {y^3} 3 +y-\frac 56}$$