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

Question

Let $y=y(x)$ be the solution of the differential equation :

$\frac{dy}{dx}+(\frac{6{x}^{2}+(3{x}^{2}+2{x}^{3}+4){e}^{-2x}}{({x}^{3}+2)(2+{e}^{-2x})})y=2+{e}^{-2x}$

$x\in (-1,2)$, satisfying $y(0)=\frac{3}{2}$. If $y(1)=\alpha (2+{e}^{-2})$, then $\alpha $ is equal to :

Accepted Answer

Given differential equation is

$\frac{dy}{dx}+(\frac{6{x}^{2}+(3{x}^{2}+2{x}^{3}+4){e}^{-2x}}{({x}^{3}+2)(2+{e}^{-2x})})y=2+{e}^{-2x}$

with initial condition

$y(0)=\frac{3}{2}.$

We have to find $y(1)=\alpha (2+{e}^{-2})$.

Compare with linear differential equation: This is of the form

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

where

$P(x)=\frac{6{x}^{2}+(3{x}^{2}+2{x}^{3}+4){e}^{-2x}}{({x}^{3}+2)(2+{e}^{-2x})},\, Q(x)=2+{e}^{-2x}.$

To solve, we find the integrating factor:

$={e}^{\int P(x)\, dx}.$

Simplify $P(x)$: Notice that

$\frac{d}{dx}({x}^{3}+2)=3{x}^{2},$

and

$\frac{d}{dx}(2+{e}^{-2x})=-2{e}^{-2x}.$

Now consider

$\frac{d}{dx}[({x}^{3}+2)(2+{e}^{-2x})].$

Using product rule,

$=(3{x}^{2})(2+{e}^{-2x})+({x}^{3}+2)(-2{e}^{-2x})$

$=6{x}^{2}+3{x}^{2}{e}^{-2x}-2{x}^{3}{e}^{-2x}-4{e}^{-2x}.$

This does not match directly. So let us try another observation.

Write numerator as

$6{x}^{2}+(3{x}^{2}+2{x}^{3}+4){e}^{-2x}.$

Now check

$\frac{d}{dx}[({x}^{3}+2)(2+{e}^{-2x})]=6{x}^{2}+(3{x}^{2}-2{x}^{3}-4){e}^{-2x}.$

Hence if we take

$\frac{d}{dx}[({x}^{3}+2)(2+{e}^{-2x}{)}^{-1}]$

it becomes complicated. So instead, let us look for a useful factorization.

Observe that

$6{x}^{2}+(3{x}^{2}+2{x}^{3}+4){e}^{-2x}=3{x}^{2}(2+{e}^{-2x})+({x}^{3}+2)\cdot 2{e}^{-2x}.$

Therefore

$P(x)=\frac{3{x}^{2}(2+{e}^{-2x})+({x}^{3}+2)\cdot 2{e}^{-2x}}{({x}^{3}+2)(2+{e}^{-2x})}$

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

So,

$P(x)=\frac{3{x}^{2}}{{x}^{3}+2}+\frac{2{e}^{-2x}}{2+{e}^{-2x}}.$

Find integrating factor: Now

$\int P(x)\, dx=\int \frac{3{x}^{2}}{{x}^{3}+2}\, dx+\int \frac{2{e}^{-2x}}{2+{e}^{-2x}}\, dx.$

For the first integral:

Let $u={x}^{3}+2$, then $du=3{x}^{2}dx$.

So,

$\int \frac{3{x}^{2}}{{x}^{3}+2}\, dx=\ln ({x}^{3}+2).$

For the second integral, note that

$\frac{d}{dx}(2+{e}^{-2x})=-2{e}^{-2x}.$

Hence

$\int \frac{2{e}^{-2x}}{2+{e}^{-2x}}\, dx=-\ln (2+{e}^{-2x}).$

Therefore,

$\int P(x)\, dx=\ln ({x}^{3}+2)-\ln (2+{e}^{-2x})=\ln (\frac{{x}^{3}+2}{2+{e}^{-2x}}).$

So the integrating factor is

$={e}^{\int P(x)\, dx}=\frac{{x}^{3}+2}{2+{e}^{-2x}}.$

Multiply the equation by the integrating factor: Multiplying throughout by

$\frac{{x}^{3}+2}{2+{e}^{-2x}},$

we get

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

So the left side becomes

$\frac{d}{dx}[y\cdot \frac{{x}^{3}+2}{2+{e}^{-2x}}].$

Hence

$\frac{d}{dx}[y\cdot \frac{{x}^{3}+2}{2+{e}^{-2x}}]={x}^{3}+2.$

Integrating both sides,

$y\cdot \frac{{x}^{3}+2}{2+{e}^{-2x}}=\int ({x}^{3}+2)\, dx=\frac{{x}^{4}}{4}+2x+C.$

Thus,

$y=\frac{2+{e}^{-2x}}{{x}^{3}+2}(\frac{{x}^{4}}{4}+2x+C).$

Use initial condition: Given

$y(0)=\frac{3}{2}.$

Substitute $x=0$:

$\frac{3}{2}=\frac{2+{e}^{0}}{{0}^{3}+2}(0+0+C)=\frac{3}{2}C.$

So,

$C=1.$

Hence solution is

$y=\frac{2+{e}^{-2x}}{{x}^{3}+2}(\frac{{x}^{4}}{4}+2x+1).$

Find $y(1)$: Substitute $x=1$:

$y(1)=\frac{2+{e}^{-2}}{{1}^{3}+2}(\frac{1}{4}+2+1).$

$y(1)=\frac{2+{e}^{-2}}{3}(\frac{1}{4}+3)=\frac{2+{e}^{-2}}{3}\cdot \frac{13}{4}$

$y(1)=\frac{13}{12}(2+{e}^{-2}).$

Given that

$y(1)=\alpha (2+{e}^{-2}),$

so

$\alpha =\frac{13}{12}.$

Final Answer
$\frac{13}{12}$

So, the correct option is D.

Solutions · 1