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Jazz Frenia · Accepted Answer

We are given the differential equation:4d2ydt2+6dydt+5y=twith initial conditions dydt(0)=y(0)=0.To find the Laplace transform of y(t), we can apply the Laplace transform to both sides of the differential equation:4ℒ{d2ydt2}+6ℒ{dydt}+5ℒ{y}=ℒ{t}Using the property of the Laplace transform that ℒ{d2ydt2}=s2ℒ{y}−s·y(0)−dydt(0) and ℒ{dydt}=sℒ{y}−y(0), we can rewrite the equation as:4s2Y(s)−4sy(0)−4dydt(0)+6sY(s)−6y(0)+5Y(s)=1s2Since dydt(0)=ddt(y(0))=0 and y(0)=0, we can simplify the equation to:4s2Y(s)+6sY(s)+5Y(s)=1s2Simplifying further, we get:(4s2+6s+5)Y(s)=1s2Therefore, the Laplace transform of y(t) is:Y(s)=1s2(4s2+6s+5)To solve for Y(s), we can use partial fraction decomposition. First, we factor the denominator:4s2+6s+5=(2s+1)2+4Thus, we can write:1s2(4s2+6s+5)=As+Bs2+Cs+D2s+1+4−Cs+D2s+1−4Simplifying, we get:Y(s)=1s2(As+Bs2+Cs+D2s+1+2−Cs+D2s+1−2)Y(s)=As+Bs4+Cs+D4s2+4s=As+Bs4+Cs+D4s(s+1)Using the initial condition y(0)=0, we know that Y(0)=0. Thus, we can solve for B:Y(0)=B04+D4(0)(0+1)=0⇒B=0Using the initial condition dydt(0)=0, we know that Y′(0)=0. Thus, we can solve for A:Y′(s)=−2As3−4As+Cs2+D4s2(s+1)Y′(0)=−0−0+C(0)+D4(0)(0+1)=0⇒D=0Now we can write Y(s) in terms of A and C:Y(s)=Ass4+C4(s+1)To solve for A and C, we can take the Laplace inverse of Y(s) using tables or the method of partial fraction decomposition:ℒ−1{Y(s)}=ℒ−1{Ass4+C4(s+1)}y(t)=at3+C4e−twhere a and C are constants that we still need to solve for.Using the initial condition dydt(0)=0, we can solve for a:dydt(t)=3at2−C4e−tdydt(0)=0=3a(0)2−C4e−(0)⇒C=0Thus, the solution to the differential equation with initial conditions is:y(t)=at3Using the initial condition y(0)=0, we get:y(0)=0=a(0)3⇒a=0Therefore, the solution to the differential equation with initial conditions is y(t)=0.In terms of Laplace transform, we have:Y(s)=1s2(4s2+6s+5)=Cs+D4s(s+1)=0s4where C=0 and D=0.

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