Question

For the following DE, check that \(y_1\) is a solution of the corresponding homogeneous equation and then calculate the general solution by the reduction of order method.

\[y^{\prime\prime} -4y^\prime +3y=3x-4;\qquad y_1 =e^x\]

Answer 1

\[y^{\prime\prime} -4y^\prime +3y=3x-4;\qquad y_1 =e^x\]
\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^\prime =e^x\]\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^{\prime\prime} =e^x\]
\[y_1^{\prime\prime} -4y_1^\prime +3y_1=0\]\[\left(e^x\right) -4\left(e^x\right)+3\left(e^x\right) =0\]\[\left(1-4+3\right)e^x=0\]
\[\therefore\qquad y_1= e^x\qquad\text{is a solution to the homogenous equation}\]

Answer 2

\[y=y_p+y_c\]\[y_p=y_1+y_2\]\[y_2=uy_1\]

Answer 3

\[y_2=uy_1=ue^x\]\[y_2^\prime=u^\prime e^x+ue^x=(u^\prime+u)e^x\]\[y_2^{\prime\prime}=(u^{\prime\prime}+u^\prime)e^x+(u^\prime+u)e^x\]\[y_2^{\prime\prime}=(u^{\prime\prime}+2u^\prime+u)e^x\]

\[y_2^{\prime\prime} -4y_2^\prime +3y_2=0\]\[\left((u^{\prime\prime}+2u^\prime+u)e^x\right) -4\left((u^\prime+u)e^x\right) +3\left(ue^x\right)=0\]\[\left(u^{\prime\prime}+2u^\prime+u -4u^\prime-4u +3u\right)e^x=0\]\[\left(u^{\prime\prime}-2u^\prime\right)e^x=0\]\[u^{\prime\prime}-2u^\prime=0\]

Answer 4

\[\text{let } p=u^\prime\]\[p^\prime=u^{\prime\prime}\]
\[p^\prime-2p=0\]

Answer 5

\[\mu(x)=e^{\int-2\text dx}=e^{-2x}\]
\[\left(pe^{-2x}\right)^\prime=0\]\[pe^{-2x}=\int 0\cdot\text dx\]\[pe^{-2x}=1\]\[p=e^{2x}\]\[u^{\prime}=e^{2x}\]\[u=\frac {e^{2x}}2\]

Answer 6

\[y_2=e^{2x}y_1\]
\[y_p=ae^x+be^xe^{2x}\]\[=ae^x+be^{3x}\]

Answer 7

how should i find \(y_c\)

Answer 8

Laplace transform is the best way to solve it.
follow me.

Answer 9

the question is asking for reduction of order method.

Answer 10

ok I will give you a beautiful solution to reduction.
follow me.

Answer 11

@mahmit2012 doesn't Laplace transform requite initial values??

Answer 12

ok