We already solved this differential equation in Example “Solving Second-Order Equations with Constant Coefficients” part a. and found the general solution to be

[latex]y(x)=c_1e^{-4x}+c_2e^x[/latex].

Then

[latex]y^\prime=-4c_1e^{-4x}+c_2e^x[/latex].

When [latex]x=0[/latex], we have [latex]y(0)=c_1+c_2[/latex] and [latex]y^\prime(0)=-4c_1+c_2[/latex]. Applying the initial conditions, we have

[latex]\begin{aligned} c_1+c_2&=1 \\ -4c_1+c_2&=-9 \end{aligned}[/latex].

Then [latex]c_1=1-c_2[/latex]. Substituting this expression into the second equation, we see that

[latex]\begin{aligned} -4(1-c_2)+c_2&=-9 \\ -4+4c_2+c_2&=-9 \\ 5c_2&=-5 \\ c_2&=-1 \end{aligned}[/latex].

So, [latex]c_1=2[/latex] and the solution to the initial-value problem is

[latex]y(x)=2e^{-4x}-e^x[/latex].

We already solved this differential equation in Example “Solving Second-Order Equations with Constant Coefficients” part b. and found the general solution to be

[latex]y(x)=e^{-3x}(c_1\cos2x+c_2\sin2x)[/latex].

Then

[latex]y^\prime(x)=e^{-3x}(-2c_1\sin2x+2c_2\cos2x)-3e^{-3x}(c_1\cos2x+c_2\sin2x)[/latex].

When [latex]x=0[/latex], we have [latex]y(0)=c_1[/latex] and [latex]y^\prime(0)=2c_2-3c_1[/latex]. Applying the initial conditions, we obtain

[latex]\begin{aligned} c_1&=0 \\ -3c_1+2c_2&= 2 \end{aligned}[/latex].

Therefore, [latex]c_1=0[/latex], [latex]c_2=1[/latex], and the solution to the initial value problem is shown in the following graph.

[latex]y=e^{-3x}\sin2x[/latex]

Figure 1.

In Example “Solving Second-Order Equations with Constant Coefficients” part c. we found the general solution to this differential equation to be

[latex]y(t)=c_1e^{-t}+c_2te^{-t}[/latex].

Then

[latex]y^\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[/latex].

When [latex]t=0[/latex], we have [latex]y(0)=c_1[/latex] and [latex]y'(0)=-c_1+c_2[/latex]. Applying the initial conditions, we obtain

[latex]\begin{aligned} c_1&=1 \\ -c_1+c_2&= 0 \end{aligned}[/latex].

Thus, [latex]c_1=1[/latex], [latex]c_2=1[/latex], and the solution to the initial value problem is

[latex]y(t)=e^{-t}+te^{-t}[/latex].

This solution is represented in the following graph. At time [latex]t=2[/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\approx0.406[/latex] [latex]m[/latex] below equilibrium.

Figure 3.

To calculate the velocity at time [latex]t=1[/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[/latex], so

[latex]y^\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[/latex].

Then [latex]y^\prime(1)=e^{-1}\approx-0.3679[/latex]. At time [latex]t=1[/latex], the mass is moving upward at [latex]0.3679[/latex] m/sec.

We have

1. Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[/latex]. So the solution is of the form [latex]y(t)=c_2\sin4t[/latex]. When we apply the second boundary condition, though, we get [latex]y\left(\frac{\pi}4\right)=c_2\sin\left(4\left(\frac{\pi}4\right)\right)=c_2\sin\pi=0[/latex] for all values of [latex]c_2[/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\sin4t[/latex] is a solution for any value of [latex]c_2[/latex].
2. Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[/latex]. Applying the second boundary condition gives[latex]y(\frac{\pi}{8})=c_2=0[/latex], so [latex]c_2=0[/latex]. In this case, we have a unique solution: [latex]y(t)=\cos 4t[/latex].
3. Applying the first boundary condition given here, we get [latex]y\left(\frac{\pi}8\right)=c_2=0[/latex]. However, applying the second boundary condition gives [latex]y\left(\frac{3\pi}8\right)=-c_2=2[/latex], so [latex]c_2=-2[/latex]. We cannot have [latex]c_2=0=-2[/latex] so this boundary value problem has no solution.