Question

(1)(16 points) (a) for the differential equation ( y^{prime prime}+3 y^{prime}=0 ) obtain a fundamental pair of solutions and compute the wronskian of the pair. (b) solve the initial value problem: ( y^{prime prime}+3 y^{prime}=2 e^{3 t}+9 ) with ( y(0)=0 ) and ( y^{prime}(0)=0 ).

Explanation:

Part (a)

Step 1: Find the characteristic equation

For the differential equation \(y'' + 3y'=0\), assume \(y = e^{rt}\). Then \(y'=re^{rt}\) and \(y'' = r^{2}e^{rt}\). Substituting into the differential equation gives \(r^{2}e^{rt}+3re^{rt}=0\), or \(e^{rt}(r^{2}+3r)=0\). Since \(e^{rt}
eq0\) for all \(t\), the characteristic equation is \(r^{2}+3r = 0\), factoring we get \(r(r + 3)=0\).

Step 2: Solve the characteristic equation

Solving \(r(r + 3)=0\) gives \(r_{1}=0\) and \(r_{2}=-3\). The general solution is \(y(t)=C_{1}e^{0t}+C_{2}e^{-3t}=C_{1}+C_{2}e^{-3t}\). A fundamental pair of solutions is \(y_{1}(t)=1\) and \(y_{2}(t)=e^{-3t}\).

Step 3: Compute the Wronskian

The Wronskian \(W(y_{1},y_{2})\) is given by the determinant \(W=

$$\begin{vmatrix}y_{1}&y_{2}\\y_{1}'&y_{2}'\end{vmatrix}$$

\). We have \(y_{1}=1,y_{1}' = 0,y_{2}=e^{-3t},y_{2}'=-3e^{-3t}\). Then \(W=

$$\begin{vmatrix}1&e^{-3t}\\0&-3e^{-3t}\end{vmatrix}$$

=1\times(- 3e^{-3t})-0\times e^{-3t}=-3e^{-3t}\)

Step 1: Find the complementary solution

The homogeneous equation \(y'' + 3y'=0\) has the characteristic equation \(r^{2}+3r=0\) with roots \(r_{1}=0\) and \(r_{2}=-3\). So the complementary solution \(y_{c}(t)=C_{1}+C_{2}e^{-3t}\)

Step 2: Find a particular solution

For the non - homogeneous term \(2e^{3t}+9\).
For \(f_{1}(t)=2e^{3t}\), assume a particular solution \(y_{p1}(t)=Ae^{3t}\). Then \(y_{p1}'=3Ae^{3t}\) and \(y_{p1}'' = 9Ae^{3t}\). Substituting into \(y'' + 3y'=2e^{3t}\) gives \(9Ae^{3t}+3\times(3Ae^{3t})=2e^{3t}\), \(9Ae^{3t}+9Ae^{3t}=2e^{3t}\), \(18Ae^{3t}=2e^{3t}\), so \(A=\frac{1}{9}\)
For \(f_{2}(t)=9\), assume a particular solution \(y_{p2}(t)=Bt\). Then \(y_{p2}'=B\) and \(y_{p2}'' = 0\). Substituting into \(y'' + 3y'=9\) gives \(0 + 3B=9\), so \(B = 3\)
The particular solution \(y_{p}(t)=\frac{1}{9}e^{3t}+3t\)

Step 3: Find the general solution

The general solution \(y(t)=y_{c}(t)+y_{p}(t)=C_{1}+C_{2}e^{-3t}+\frac{1}{9}e^{3t}+3t\)

Step 4: Apply the initial conditions

\(y(0)=0\): \(C_{1}+C_{2}+\frac{1}{9}=0\)
\(y'(t)=-3C_{2}e^{-3t}+\frac{1}{3}e^{3t}+3\), \(y'(0)=0\): \(-3C_{2}+\frac{1}{3}+3 = 0\), \(-3C_{2}=-\frac{10}{3}\), \(C_{2}=\frac{10}{9}\)
Substitute \(C_{2}=\frac{10}{9}\) into \(C_{1}+C_{2}+\frac{1}{9}=0\), we get \(C_{1}=- \frac{11}{9}\)

Answer:

A fundamental pair of solutions is \(y_{1}(t) = 1\) and \(y_{2}(t)=e^{-3t}\), and the Wronskian \(W=-3e^{-3t}\)

Part (b)