Solve the initial value problem x" + x = 8 sin t, x(0) = 3, x'(0) = 1.

Question

Solve the initial value problem

[latex]x^{\prime \prime}+x=8 \sin t, \quad x(0)=3, \quad x^{\prime}(0)=1[/latex].

Accepted Answer

Using the substitution [latex]x_{1}=x, x_{2}=x^{\prime}[/latex], we can write this in matrix form:

[latex]\left(\begin{array}{l}x_{1} \x_{2}\end{array} ight)^{\prime}=\left(\begin{array}{rr}0 & 1 \-1 & 0\end{array} ight)\left(\begin{array}{l}x_{1} \x_{2}\end{array} ight)+\left(\begin{array}{c}0 \8 \sin t\end{array} ight)[/latex]

The associated homogeneous system has the principal matrix solution

[latex]\Psi(t)=\left(\begin{array}{rr}\cos t & \sin t \-\sin t & \cos t\end{array} ight)[/latex]

Then

[latex]\Psi^{-1}(t)=\left(\begin{array}{rr}\cos t & -\sin t \\sin t & \cos t\end{array} ight)[/latex]

and

[latex]\begin{aligned}\int_{0}^{t} \Psi^{-1}(s) \mathbf{f}(s) d s &=\int_{0}^{t}\left(\begin{array}{cc}\cos s & -\sin s \\sin s & \cos s\end{array} ight)\left(\begin{array}{c}0 \8 \sin s\end{array} ight) d s \&=\int_{0}^{t}\left(\begin{array}{c}-8 \sin ^{2} s \8 \sin s \cos s\end{array} ight) d s=\left(\begin{array}{c}-4 t+4 \sin t \cos t \4 \sin ^{2} t\end{array} ight)\end{aligned}[/latex]

Since the solution [latex]\varphi(t)[/latex] satisfies the initial conditions

[latex]\varphi(0)=\left(\begin{array}{l}3 \1\end{array} ight)[/latex]

we obtain from equation (14)

[latex]\varphi(t)=\Psi(t) \mathbf{x}_{0}+\Psi(t) \int_{t_{0}}^{t} \Psi^{-1}(s) \mathbf{f}(s) d s[/latex]

[latex]\begin{aligned}\varphi(t) &=\left(\begin{array}{rr}\cos t & \sin t \-\sin t & \cos t\end{array} ight)\left(\begin{array}{l}3 \1\end{array} ight)+\left(\begin{array}{rc}\cos t & \sin t \-\sin t & \cos t\end{array} ight)\left(\begin{array}{c}-4 t+4 \sin t \cos t \4 \sin ^{2} t\end{array} ight) \&=\left(\begin{array}{c}-4 t \cos t+5 \sin t+3 \cos t \4 t \sin t-3 \sin t+\cos t\end{array} ight)\end{aligned}[/latex]

Here we have used the identity

[latex]\sin ^{3} t+\cos ^{2} t \sin t=\sin t\left(\sin ^{2} t+\cos ^{2} t ight)=\sin t[/latex].

Thus, [latex]x(t)=-4 t \cos t+5 \sin t+3 \cos t[/latex] is the solution to the problem.

Question 12.7.2: Solve the initial value problem x" + x = 8 sin t, x(0) = 3, ...

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