Both equations are linear, so I'll use the integrating factor method.
The first ODE
[tex]xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0[/tex]
has integrating factor
[tex]\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x[/tex]
In the original equation, multiply both sides by eˣ :
[tex]xe^x y' + (x+1) e^x y = 0[/tex]
Observe that
d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ
so that the left side is the derivative of a product, namely
[tex]\left(xe^xy\right)' = 0[/tex]
Integrate both sides with respect to x :
[tex]\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx[/tex]
[tex]xe^xy = C[/tex]
Solve for y :
[tex]y = \dfrac{C}{xe^x}[/tex]
Use the given initial condition to solve for C. When x = 1, y = 2, so
[tex]2 = \dfrac{C}{1\cdot e^1} \implies C = 2e[/tex]
Then the particular solution is
[tex]\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}[/tex]
The second ODE
[tex](1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0[/tex]
has integrating factor
[tex]\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}[/tex]
Multiply both sides of the equation by 1/(1 + x²) :
[tex]\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0[/tex]
and observe that
d/dx[1/(1 + x²)] = -2x/(1 + x²)²
Then
[tex]\left(\dfrac1{1+x^2}y\right)' = 0[/tex]
[tex]\dfrac1{1+x^2}y = C[/tex]
[tex]y = C(1 + x^2)[/tex]
When x = 0, y = 3, so
[tex]3 = C(1+0^2) \implies C=3[/tex]
[tex]\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}[/tex]
