Question

(a). Find the general solution of the DE x.dy/dx - y= x^2e^-x, giving your answer in the form y=f(x). (b) Verify that the graphs of all solutions of the DE pass through the origin O, and find the particular solution which is such that dy/dx=-1 at O

Answer 1

$$x\frac{dy}{dx}-y=x^2e^{-x}$$Divide through by \(x\) on both sides to get rid of our coefficient of \(\frac{dy}{dx}\).$$\frac{dy}{dx}-\frac1xy=xe^{-x}$$Hey, this is a linear differential equation. We want our left-hand side to resemble something we can use the product rule to simplify so we need to multiply by an *integration factor* \(\mu\); in this case$$\mu=e^{\int-\frac1xdx}=e^{-\int\frac1x dx}=e^{-\log x}=\frac1x$$Multiply both sides by \(\mu=\frac1x\) to yield:$$\frac1x\frac{dy}{dx}-\frac1{x^2}y=e^{-x}$$Recognize that our left-hand side is just:$$\frac1x\frac{dy}{dx}+\left(-\frac1{x^2}\right)y=\frac{d}{dx}\left(\frac1xy\right)$$So we integrate both sides:$$\frac{d}{dx}\left(\frac1xy\right)=e^{-x}\\\int\frac{d}{dx}\left(\frac1xy\right)dx=\int e^{-x}dx\\\frac1xy=-e^{-x}+C\\y=-xe^{-x}+Cx$$

Answer 2

Now to verify that all solutions pass through the origin \((0,0)\), we just check to see if \(y(0)=0\).$$\begin{align*}y&=-xe^{-x}+Cx\\0&=-(0)e^{-(0)}+C(0)\\&=0+0\\&=0\end{align*}$$
Indeed, they all pass through \((0,0)\) -- regardless of \(C\).

Answer 3

thank you soooo much! this helped alot!

Answer 4

For our particular solution given the initial condition \(\frac{dy}{dx}=-1\) at \(x=0\), recognize our derivative is merely:$$\begin{align*}\frac{dy}{dx}&=\frac{d}{dx}(-xe^{-x}+Cx)\\&=-\frac{d}{dx}xe^{-x}+C\\&=-(e^{-x}-xe^{-x})+C\\&=-e^{-x}+xe^{-x}+C\end{align*}$$Since we're interested in \(\frac{dy}{dx}=-1\), we substitute and reduce:$$\begin{align*}-1&=-e^{-(0)}+0e^{-(0)}+C\\&=-1+0+C\\&=-1+C\\0&=C\end{align*}$$

Answer 5

i now see my error, thank you

Answer 6

So our particular solution is then \(y=-xe^{-x}+0=-xe^{-x}\).

Answer 7

np @sellieG