Question #122598
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) = _____
1
Expert's answer
2020-07-05T17:33:17-0400

1.The given equation can be written as (D28D+16)y=12x+6(D^2 − 8D + 16)y = 12x + 6.

The auxiliary equation is, m28m+16=0.m^2 − 8m + 16 = 0. Solving we get, m=4(twice)m = 4 (twice) . Thus, the complementary function is, yc=(c1x+c2)e4xy_{c}=(c_{1}x+c_{2})e^{4x}. Let yp=Ax+By_{p} = Ax+B be the trial particular solution. Then,


yp8yp+16yp=12x+6 08A+16(Ax+B)=12x+6 A=34,B=34 Hence, yp=34(x+1)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=yc+yp=(c1x+c2)e4x+34(x+1).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 (D2+7)y=7x.(D^{2}+7)y=7x. The auxiliary equation is, m2+7=0.m^{2}+7=0. Solving we get, m=±i7m=\pm i\sqrt{7}. Thus, the complementary function is, yc=c1cos(7x)+c2sin(7x).y_{c}=c_{1}\cos(\sqrt{7}x)+c_{2}\sin(\sqrt{7}x). The particular solution is


yp=1D2+77x        =11+D27x        =(1+D27)1x        =xy_{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=c1cos(7x)+c2sin(7x)+xy = 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.y(0)=0; ~y(1)+y'(1)=0. Using these conditions we get,

c1=0 c2=27cos(7)+sin(7)c_{1}=0\\~\\ c_{2}=\dfrac{-2}{\sqrt{7}\cos(\sqrt7)+\sin(\sqrt{7})}


The general solution is,


y=x2sin(7x)7cos(7)+sin(7)y = x-\dfrac{2\sin(\sqrt{7}x)}{\sqrt{7}\cos(\sqrt{7})+\sin(\sqrt{7})}


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