Question #324463

Determine the unique solution of the following differential equations by using Laplace transforms:

y''(t) + 2y'(t) -3y(t) = e-3t if y'(0) = 0 and y(0) = 0


1
Expert's answer
2022-04-14T09:22:55-0400

y(t)+2y(t)3y(t)=e3t,y(0)=0,y(0)=0y(t)Y(p)p2Y(p)py(0)y(0)+2(pY(p)y(0))3Y(p)=1p+3(p2+2p3)Y(p)=1p+3Y(p)=1(p+3)2(p1)1(p+3)2(p1)=A(p+3)2+Bp+3+Cp11=A(p1)+B(p+3)(p1)+C(p+3)2(B+C)p2+(A+2B+6C)p+(A3B+9C)=1{B+C=0A+2B+6C=0A3B+9C=1A=14,B=116,C=116Y(p)=14(p+3)2116(p+3)+116(p1)14te3t116e3t+116ety(t)=14te3t116e3t+116ety''\left( t \right) +2y'\left( t \right) -3y\left( t \right) =e^{-3t},y'\left( 0 \right) =0,y\left( 0 \right) =0\\y\left( t \right) \rightarrow Y\left( p \right) \\p^2Y\left( p \right) -py\left( 0 \right) -y'\left( 0 \right) +2\left( pY\left( p \right) -y\left( 0 \right) \right) -3Y\left( p \right) =\frac{1}{p+3}\\\left( p^2+2p-3 \right) Y\left( p \right) =\frac{1}{p+3}\\Y\left( p \right) =\frac{1}{\left( p+3 \right) ^2\left( p-1 \right)}\\\frac{1}{\left( p+3 \right) ^2\left( p-1 \right)}=\frac{A}{\left( p+3 \right) ^2}+\frac{B}{p+3}+\frac{C}{p-1}\\1=A\left( p-1 \right) +B\left( p+3 \right) \left( p-1 \right) +C\left( p+3 \right) ^2\\\left( B+C \right) p^2+\left( A+2B+6C \right) p+\left( -A-3B+9C \right) =1\\\left\{ \begin{array}{c} B+C=0\\ A+2B+6C=0\\ -A-3B+9C=1\\\end{array} \right. \Rightarrow A=-\frac{1}{4},B=-\frac{1}{16},C=\frac{1}{16}\\Y\left( p \right) =-\frac{1}{4\left( p+3 \right) ^2}-\frac{1}{16\left( p+3 \right)}+\frac{1}{16\left( p-1 \right)}\mapsto -\frac{1}{4}te^{-3t}-\frac{1}{16}e^{-3t}+\frac{1}{16}e^t\\y\left( t \right) =-\frac{1}{4}te^{-3t}-\frac{1}{16}e^{-3t}+\frac{1}{16}e^t


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS