Second Order ODEs Practice

The highest derivative is \(y''\), so it is a second-order ODE.

The equation is linear because \(y\), \(y'\), and \(y''\) each appear to the first power and are not multiplied together.

It is homogeneous because the right-hand side is \(0\).

It has constant coefficients because \(1\), \(3\), and \(-4\) are constants.

Divide every term by \(4\):

So the normalized form is

Write the equation as

The function \(f\) is continuous everywhere, and its partial derivatives with respect to \(y\) and \(y'\) are also continuous everywhere.

Therefore the theorem guarantees a unique local solution near \(x=0\).

Try \(y=e^{rx}\). The characteristic equation is

Factor:

So the roots are \(r=2\) and \(r=3\). The general solution is

Try \(y=e^{rx}\). The characteristic equation is

This is a repeated root \(r=2\), so

Try \(y=e^{rx}\). The characteristic equation is

so

The general real solution is

The homogeneous characteristic equation is

so \(e^x\) and \(xe^x\) already appear in the homogeneous solution.

To avoid duplication, multiply by \(x\) again. A correct trial is

Use the Euler-Cauchy trial \(y=x^m\). Then

Substitute:

which simplifies to

So

and \(m=\pm 2\). Therefore

Use

Here,

So

On any interval that does not include \(0\), the functions are linearly independent.

Compute the natural frequency:

Compute the damping ratio:

Since \(\zeta=1\), the system is critically damped.

The characteristic equation is

So

Use \(y(0)=1\):

Differentiate:

Use \(y'(0)=4\):

Subtract the first equation from the second:

Then \(C_1 = -2\). So

The homogeneous equation has characteristic equation

so

Because the forcing is \(e^x\), use a resonant trial

Substituting gives \(A=\tfrac12\), so

Now apply the initial conditions. From \(y(0)=0\),

Write

Then

Using \(y'(0)=1\) and \(C_1=0\),

So the solution is

Use the Euler-Cauchy trial \(y=x^m\). The auxiliary equation is

so \(m=\pm 2\). Thus

Apply \(y(1)=3\):

Differentiate:

Apply \(y'(1)=-1\):

Solve the system to get

Therefore

This is an Euler-Cauchy equation. Using the known solution \(y_1=x\), reduction of order gives a second independent solution \(y_2=x^2\).

So the general solution is

The characteristic equation is

so the general solution is

Use \(y(0)=0\):

Then

Now use \(y(1)=0\):

which is true for every \(C_2\).

So the boundary value problem has infinitely many solutions:

The characteristic equation is

This is a repeated root, so

Use \(y(0)=2\):

Differentiate:

Use \(y'(0)=0\):

so

Thus

Because the root is repeated, the motion is critically damped.

Since the forcing is \(\sin(2t)\) and there is no resonance, try

Then

Substitute into the equation:

Match coefficients with \(6\sin(2t)\):

So \(A=0\) and \(B=\frac{6}{5}\). A steady-state response is

In normalized form, the coefficients are

The function \(p(x)\) is undefined at \(x=1\), and \(q(x)=\ln x\) requires \(x>0\).

So the coefficients are continuous on

That is the largest open interval containing \(x=2\) on which a unique local solution is guaranteed.

For an undamped spring, the equation is

Here,

so

The general solution is

Use \(y(0)=1/2\):

Released from rest means \(y'(0)=0\). Since

we get \(C_2=0\).

So

The first return to equilibrium happens when \(y(t)=0\), so

Thus the first return time is

The characteristic equation is

So

Apply the initial conditions:

Differentiate:

Then

so \(C_2=8\).

Therefore

Try

Then

Match coefficients with \(6\sin(2t)\):

So a particular solution is

The general solution is

Use \(y(0)=0\):

So

Now apply \(y\left(\frac12\right)=0\):

Thus the only solution is

The characteristic equation is

So

The \(e^{2x}\) term grows as \(x\to\infty\), while the \(e^{-x}\) term decays.

So a generic solution grows without bound, and the equilibrium is unstable unless \(C_1=0\).

The homogeneous equation has characteristic roots \(\pm 2i\), so

Because the forcing is \(\cos(2x)\), which resonates with the homogeneous solution, try

For this choice,

Match coefficients with \(8\cos(2x)\):

So

Apply the initial conditions:

and

Therefore

The characteristic equation is

so

Thus

For the forcing term \(10e^{-x}\), try

Substitute:

Set this equal to \(10e^{-x}\):

So

Use \(y(0)=1\):

Differentiate in the form \(y=e^{-x}u(x)\), where

Then

At \(x=0\),

Using \(y'(0)=0\),

so

Therefore

Use the Euler-Cauchy trial \(y=x^m\). The auxiliary equation is

which simplifies to

So \(m=\pm 1\), and

Apply \(y(1)=2\):

Apply \(y(2)=3\):

Solve the system:

So the solution is

The homogeneous equation \(y''+y=0\) has fundamental solutions

Their Wronskian is

For variation of parameters,

so

Also,

so

Thus a particular solution is

Therefore the general solution is