Interpret the differential equation

The given equation is

$(x+1)\,dy = \big((x+1)^2 + y - 3\big)\,dx$

with condition

$y(2)=0.$

So,

$\frac{dy}{dx} = \frac{(x+1)^2 + y - 3}{x+1}.$

Simplify the right-hand side

Expand:

$(x+1)^2 - 3 = x^2+2x+1-3 = x^2+2x-2.$

Hence,

$\frac{dy}{dx} = \frac{x^2+2x-2+y}{x+1}.$

Write it as

$\frac{dy}{dx} - \frac{1}{x+1}y = \frac{x^2+2x-2}{x+1}.$

Simplify the non-homogeneous term

Divide:

$\frac{x^2+2x-2}{x+1} = x+1 - \frac{3}{x+1}.$

Therefore the differential equation becomes

$\frac{dy}{dx} - \frac{1}{x+1}y = x+1 - \frac{3}{x+1}.$

Solve as a linear differential equation

Standard form:

$\frac{dy}{dx} + P(x)y = Q(x),$

where

$P(x) = -\frac{1}{x+1}, \qquad Q(x)=x+1-\frac{3}{x+1}.$

The integrating factor is

$\text{I.F.} = e^{\int -\frac{1}{x+1}\,dx} = e^{-\ln(x+1)} = \frac{1}{x+1}$

since $x\ge 0$, so $x+1>0$.

Multiply through by the integrating factor

Multiplying the equation by $\dfrac{1}{x+1}$,

$\frac{1}{x+1}\frac{dy}{dx} - \frac{y}{(x+1)^2} = 1 - \frac{3}{(x+1)^2}.$

The left side is

$\frac{d}{dx}\left(\frac{y}{x+1}\right).$

So,

$\frac{d}{dx}\left(\frac{y}{x+1}\right) = 1 - \frac{3}{(x+1)^2}.$

Integrate

$\frac{y}{x+1} = \int \left(1 - \frac{3}{(x+1)^2}\right)dx.$

Now,

$\int 1\,dx = x,$

and

$\int -\frac{3}{(x+1)^2}\,dx = -3\int (x+1)^{-2}dx = -3\left(-(x+1)^{-1}\right)=\frac{3}{x+1}.$

Hence,

$\frac{y}{x+1} = x + \frac{3}{x+1} + C.$

Multiply by $(x+1)$:

$y = x(x+1) + 3 + C(x+1).$

So,

$y = x^2+x+3 + C(x+1).$

Use the initial condition $y(2)=0$

Substitute $x=2$, $y=0$:

$0 = 2^2+2+3 + C(3) = 9 + 3C.$

Therefore,

$C=-3.$

Thus,

$y = x^2+x+3 - 3(x+1) = x^2-2x.$

Find $y(3)$

$y(3)=3^2-2\cdot 3=9-6=3.$

Compare with stored answer

Derived answer is $3$, which matches the stored correct answer.