Find the continuous solution to the initial value problem

Question

Find the continuous solution satisfying $(1+x^2)y'+2xy=f(x),f(x)=\left\{\begin{matrix} x& 0\leq x < 1\\ -x& x\geq 1 \end{matrix}\right.$

with y(0)=0

My attempt:

$(1+x^2)y'+2xy=f(x)\Rightarrow ((1+x^2)y)'=f(x)$

$(1+x^2)y=\int f(x)dx=\int ^1_0 x dx-\int^\infty_1 x dx$

i get $\infty$ R.H.S is this process is right

Answer 1

No. Since $y(0)=0$, we have

$(1+x^2)y=\int_0^x f(t)dt$.

For $0 \le x \le 1$ we get

$(1+x^2)y=\int_0^x t dt=\frac{1}{2}x^2$

and for $x>1$ we derive

$(1+x^2)y=\int_0^1 t dt+\int_1^x (-t) dt=1-\frac{1}{2}x^2$.