Answer to Question #122598 in Differential Equations for jse

Question

1. Solve the given differential equation by undetermined coefficients.
y'' − 8y' + 16y = 12x + 6

y(x) =_____

2. Solve the given boundary-value problem.
y'' + 7y = 7x, y(0) = 0, y(1) + y'(1) = 0

y(x) = _____

Answer 1

1.The given equation can be written as $(D^2 − 8D + 16)y = 12x + 6$.

The auxiliary equation is, $m^2 − 8m + 16 = 0.$ Solving we get, $m = 4 (twice)$ . Thus, the complementary function is, $y_{c}=(c_{1}x+c_{2})e^{4x}$. Let $y_{p} = Ax+B$ be the trial particular solution. Then,

$y_{p}'' − 8y_{p}' + 16y_{p} = 12x + 6\\~\\ 0-8A+16(Ax+B)=12x+6\\~\\ \therefore A= \frac{3}{4}, B= \frac{3}{4}\\~\\ \text{Hence,}~ y_{p}=\frac{3}{4}(x+1)$

Thus, the general solution is, $y=y_{c}+y_{p} = (c_{1}x+c_{2})e^{4x}+\frac{3}{4}(x+1).$

2.The given equation can be written as $(D^{2}+7)y=7x.$ The auxiliary equation is, $m^{2}+7=0.$ Solving we get, $m=\pm i\sqrt{7}$. Thus, the complementary function is, $y_{c}=c_{1}\cos(\sqrt{7}x)+c_{2}\sin(\sqrt{7}x).$ The particular solution is

$y_{p} = \dfrac{1}{D^{2}+7}7x\\~\\ ~~~~~~~=\dfrac{1}{1+\frac{D^{2}}{7}}x\\~\\ ~~~~~~~=\bigg(1+\dfrac{D^{2}}{7}\bigg)^{-1}x\\~\\ ~~~~~~~= x$

$y = c_{1}\cos(\sqrt{7}x)+c_{2}\sin(\sqrt{7}x)+x$ is the general solution.

Given the boundary conditions, $y(0)=0; ~y(1)+y'(1)=0.$ Using these conditions we get,

$c_{1}=0\\~\\ c_{2}=\dfrac{-2}{\sqrt{7}\cos(\sqrt7)+\sin(\sqrt{7})}$

The general solution is,

$y = x-\dfrac{2\sin(\sqrt{7}x)}{\sqrt{7}\cos(\sqrt{7})+\sin(\sqrt{7})}$