Let y=y(x) be the solution of [(dy)/(dx)+(6x^2+(3x^2+2x^3+4)e^-2x(x^3+2)(2+e^-2x))y = 2+e^…

Question

Let $y=y(x)$ be the solution of
$$\frac{dy}{dx}+\left(\frac{6x^2+(3x^2+2x^3+4)e^{-2x}}{(x^3+2)(2+e^{-2x})}\right)y = 2+e^{-2x},\qquad x\in(-1,2),\quad y(0)=\frac32.$$
If $y(1)=\alpha(2+e^{-2})$, then $\alpha$ is equal to:

Accepted Answer

Observe that $P(x) = \dfrac{3x^2}{x^3+2}+\dfrac{2e^{-2x}}{2+e^{-2x}}$, so
$$\int P(x)\,dx = \ln(x^3+2)-\ln(2+e^{-2x}) = \ln\!\left(\frac{x^3+2}{2+e^{-2x}}\right),\qquad \text{I.F.}=\frac{x^3+2}{2+e^{-2x}}.$$

Multiplying through:
$$\frac{d}{dx}\left[y\cdot\frac{x^3+2}{2+e^{-2x}}\right] = x^3+2 \implies y\cdot\frac{x^3+2}{2+e^{-2x}} = \frac{x^4}{4}+2x+C.$$

Using $y(0)=3/2$: $\dfrac32 = \dfrac{2}{3}C \Rightarrow C=1$.

So $y = \dfrac{2+e^{-2x}}{x^3+2}\left(\dfrac{x^4}{4}+2x+1\right)$. At $x=1$:
$$y(1) = \frac{2+e^{-2}}{3}\left(\frac14+3\right) = \frac{2+e^{-2}}{3}\cdot\frac{13}{4} = \frac{13}{12}(2+e^{-2}).$$

So $\alpha = \dfrac{13}{12}$.

Solutions · 1