Let y=y(x) solve dydx+2x^2+11x+13x^3+6x^2+11x+6y = x+3x+1, x-1, passing through (0,1).

Question

Let $y=y(x)$ solve $\dfrac{dy}{dx}+\dfrac{2x^2+11x+13}{x^3+6x^2+11x+6}y = \dfrac{x+3}{x+1}$, $x>-1$, passing through $(0,1)$. Then $y(1)$ is:

Accepted Answer

Partial fractions: $\dfrac{2x^2+11x+13}{(x+1)(x+2)(x+3)} = \dfrac{2}{x+1}+\dfrac{1}{x+2}-\dfrac{1}{x+3}$.

I.F. $=\dfrac{(x+1)^2(x+2)}{x+3}$.
$$y\cdot\frac{(x+1)^2(x+2)}{x+3} = \int(x+1)(x+2)\,dx = \frac{x^3}{3}+\frac{3x^2}{2}+2x+c.$$

Using $(0,1)$: $\dfrac{2}{3}=c \implies c=\dfrac23$.

At $x=1$: $y\cdot\dfrac{4\cdot3}{4} = \dfrac13+\dfrac32+2+\dfrac23 \implies 3y=\dfrac{9}{2} \implies y=\dfrac32$.

Solutions · 1