# Variation of parameters

Nonhomogeneous ODEs that don't have constant coefficients or an $r(x)$ in our
table can't be solved using the
[method of undetermined coefficients](./undetermined-coefficients.md). An
alternative for such ODEs is the method of variation of parameters.

```{topic} Variation of parameters

For a nonhomogeneous linear second-order ODE $y'' + p(x)y' + q(x)y = r(x)$,
if $y_{\rm h} = c_1 y_1 + c_2 y_2$ solves the corresponding homogenous ODE
$y_{\rm h}'' + p(x) y_{\rm h}' + q(x) y_{\rm h} = 0$, then the particular
solution $y_{\rm p}$ is:

\begin{equation}
y_{\rm p} = -y_1 \int\frac{y_2 r}{W} \d{x} + y_2 \int \frac{y_1 r}{W} \d{x}
\end{equation}

where
\begin{equation}
W = \begin{vmatrix}y_1 & y_2 \\y_1' & y_2'\end{vmatrix} = y_1 y_2' - y_2 y_1'
\end{equation}

is the Wronksi determinant (or "Wronskian").
```

For example, consider the ODE

\begin{equation}
y'' + y = \sec x
\end{equation}

The secant is not in our table of right-hand sides, so we will use variation
of parameters to find a particular solution. The homogenous ODE is solved by
$y_1 = \cos x$ and $y_2 = \sin x$ so

\begin{align}
y_1 &= \cos x & y_2 &= \sin x \\
y_1' &= -\sin x & y_2' &= \cos x
\end{align}

and

\begin{equation}
W = \cos^2 x - (-\sin^2 x) = 1
\end{equation}

Hence,

\begin{align}
y_{\rm p} &= -\cos x\int\sin x\sec x\d{x} +\sin x\int\cos x\sec x\d{x} \\
&= \cos x\int \frac{-\sin x}{\cos x}\d{x} + \sin x \int \d{x} \\
&= \cos x\ln|\cos x| + x\sin x
\end{align}

and

\begin{equation}
y = c_1 \cos x + c_2 \sin x + \cos x\ln|\cos x| + x\sin x
\end{equation}

```{example} Steady heat diffusion with internal source
We are solving the steady-state heat diffusion equation with an internal source:

\begin{equation}
\alpha \dd{2}{T}{x} + qx \sin x = 0
\end{equation}

where $\alpha$ is thermal diffusivity, and $qx \sin x$ is the heat source. The
boundary conditions are:

\begin{equation}
T(-1) = T(1) = 1
\end{equation}

---

In standard form,

\begin{equation}
T'' = -q x \sin x
\end{equation}

1. Find homogeneous solution:

   \begin{align}
   \alpha T_{\rm h}'' &= 0 \\
   \alpha \lambda^2 &= 0 \\
   \lambda &= 0
   \end{align}

   so,

   \begin{equation}
   T_{\rm h} = c_1 + c_2 x
   \end{equation}

2. Find particular solution using variation of parameters:

   \begin{align}
   y_1 &= 1, \quad y_2 = x \\
   y_1' &= 0, \quad y_2' = 1
   \end{align}

   so the Wronskian is $W = 1 \cdot 1 - 0 \cdot x = 1$. The right-hand side
   function is

   \begin{equation}
   r = -\frac{qx \sin x}{\alpha}
   \end{equation}

   so the particular solution is:

   \begin{align}
   T_{\rm p} &= -1 \int x \left( \frac{-q x \sin x}{\alpha} \right) \, \d{x}
     + x \int \left( \frac{-q x \sin x}{\alpha} \right) \, \d{x} \\
   &= \frac{q}{\alpha} \int x^2 \sin x \, \d{x}
     - \frac{q}{\alpha} x \int x \sin x \, \d{x}
   \end{align}

   These integrals can both be evaluated by parts. For the first integral,
   use the tabular method:

   | sign | $u$     | $\d{v}$     |
   |------|---------|-------------|
   |      |         | $\sin x$    |
   |  $+$ | $x^2$   | $-\cos x$   |
   |  $-$ | $2x$    | $-\sin x$   |
   |  $+$ | $2$     | $\cos x$    |
   |      | $0$     |             |

   so

   \begin{equation}
   \int x^2 \sin x \, \d{x} = -x^2 \cos x + 2x \sin x + 2 \cos x
   \end{equation}

   For the second integral, use

   \begin{align}
   u &= x & \d{v} &= \sin x \d{x} \\
   \d{u} &= 1 \d{x} & v &= -\cos{x}
   \end{align}

   so

   \begin{equation}
   \int x \sin x \d{x} = -x \cos x + \int \cos x \d{x} = -x \cos x + \sin x
   \end{equation}

   Substitute everything back:

   \begin{align}
   T_{\rm p} &= \frac{q}{\alpha} \left[ -x^2 \cos x + 2x \sin x + 2 \cos x \right]
     - \frac{q}{\alpha} x \left[ -x \cos x + \sin x \right] \\
   &= \frac{q}{\alpha} [2 \cos x + x \sin x]
   \end{align}

3. Combine and apply boundary conditions.

   The general solution is

   \begin{equation}
   T = c_1 + c_2 x + \frac{q}{\alpha}(2\cos x + x \sin x)
   \end{equation}

   The boundary conditions are:

   \begin{align}

   T(-1) &= c_1 - c_2 + \frac{q}{\alpha}[2\cos(-1) - \sin(-1)] = 1\\
     &= c_1 - c_2 + \frac{q}{\alpha}[2\cos 1 + \sin 1] \\
   T(1) &= c_1 + c_2 + \frac{q}{\alpha}[2\cos 1 + \sin 1] = 1
   \end{align}

   This is a system of linear equations that can be solved for $c_1$ and $c_2$.
   Subtracting the two equations gives:

   \begin{equation}
   0 = 2 c_2
   \end{equation}

   so $c_2 = 0$. Plugging back in and solving for $c_1$:

   \begin{align}
   1 &= c_1 + \frac{q}{\alpha}(2\cos 1 + \sin 1) \\
   c_1 &= 1 - \frac{q}{\alpha}(2\cos 1 + \sin 1)
   \end{align}

The final result is:
\begin{equation}
T = 1 + \frac{q}{\alpha}\left[2(\cos x - \cos 1) + x \sin x - \sin 1\right]
\end{equation}
```

## Skill builder problems

1. Solve the differential equation

   \begin{equation}
   y'' - 4y' + 4y = x^{2} e^{x},  \quad y(0) = 0, \quad y'(0) = 0
   \end{equation}

   ```{solution}
   1. Find homogeneous solution $y_{\rm h}$:

      \begin{align}
      y_{\rm h}'' - 4y_{\rm h}' + 4y_{\rm h} &= 0 \\
      \lambda^2 - 4\lambda + 4 &= 0 \\
      (\lambda -2)^2 &= 0
      \end{align}

      so $\lambda = 2$ is a repeated eigenvalue and

      \begin{equation}
      y_{\rm h} = (c_1 + c_2 x) e^{2x}
      \end{equation}

   2. Find particular solution $y_{\rm p}$:

      \begin{align}
      y_1   &= e^{2x}          &\quad y_2&= x e^{2x} \\
      y_1'  &= 2 e^{2x}        &\quad y_2'&= (1 + 2x) e^{2x}
      \end{align}

      so the Wronskian is $W =(1 + 2x) \cdot e^{4x} - 2x \cdot e^{4x} = e^{4x}$.
      The right-hand side function is

      \begin{equation}
      r = x^{2} e^{x}
      \end{equation}

      so the particular solution is:

      \begin{align}
      y_{\rm p} &= -e^{2x} \int \frac{x e^{2x} (x^{2} e^{x})}{e^{4x}} \d{x} +
       x e^{2x} \int \frac{e^{2x} (x^{2} e^{x})}{e^{4x}} \d{x} \\
       &= -e^{2x} \int x^{3} e^{-x} \d{x} + x e^{2x} \int x^{2} e^{-x} \d{x}
      \end{align}

      These integrals can both be evaluated by parts using the tabular method:

      For the first

      | sign | $u$     | $\d{v}$     |
      |------|---------|-------------|
      |      |         | $e^{-x}$    |
      |  $+$ | $x^3$   | $-e^{-x}$   |
      |  $-$ | $3x^2$  | $e^{-x}$    |
      |  $+$ | $6x$    | $-e^{-x}$   |
      |  $-$ | $6$     | $e^{-x}$    |
      |      | $0$     |             |

      so

      \begin{equation}
      \int x^3 e^{-x} \d{x} = -x^3 e^{-x} - 3x^2 e^{-x} - 6x e^{-x}- 6 e^{-x}
      \end{equation}

      For the second

      | sign | $u$     | $\d{v}$     |
      |------|---------|-------------|
      |      |         | $e^{-x}$    |
      |  $+$ | $x^2$   | $-e^{-x}$   |
      |  $-$ | $2x$    | $e^{-x}$    |
      |  $+$ | $2$     | $-e^{-x}$   |
      |      | $0$     |             |

      so

      \begin{equation}
      \int x^2 e^{-x} \d{x} = -x^2 e^{-x} - 2x e^{-x} - 2 e^{-x}
      \end{equation}

      Substitute everything back:

      \begin{align}
      y_{\rm p} &= -e^{2x}
        \left( -x^3 e^{-x} - 3x^2 e^{-x} - 6x e^{-x} - 6 e^{-x} \right) \\
        &\quad + x e^{2x} \left( -x^2 e^{-x} - 2x e^{-x} - 2 e^{-x} \right) \\
        &= (x^2 + 4x + 6) e^x
      \end{align}

   3. Combine and apply initial conditions:

      The general solution and its first derivative are

      \begin{align}
      y &= (c_1 + c_2 x) e^{2x}+(x^{2}+4x+6)e^{x} \\
      y' &= 2(c_1 + c_2 x) e^{2x} + c_2e^{2x} + (x^{2}+4x+6) e^{x} + (2x+4)e^{x}
      \end{align}

      Plugging in intial conditions

      \begin{align}
      y(0) &= c_1 + 6 = 0
      y'(0) &= 2 c_1 + c_2 + 10 = 0
      \end{align}

      gives $c_1 = -6$ and $c_2 = 2$.

    The final solution is:

    \begin{equation}
    y = (2 x - 6) e^{2x} + (x^{2}+4x+6)e^{x}
    \end{equation}
   ```

2. Solve the differential equation

   \begin{equation}
   y'' + y = \csc x, \quad y(\pi/2), \quad y'(\pi/2) = 0
   \end{equation}

   ```{solution}
   1. Find homogeneous solution $y_{\rm h}$:

      \begin{align}
      y_{\rm h}' + y_{\rm h} &= 0 \\
      \lambda^2 + 1 &= 0
      \end{align}

      so $\lambda_{1,2} = \pm i$ and:

      \begin{equation}
      y_{\rm h} = c_1 \cos x + c_2 \sin_x
      \end{equation}

   2. Find particular solution $y_{\rm p}$:

      \begin{align}
      y_1 &= \cos x , \quad y_2 = \sin x \\
      y_1' &= -\sin x, \quad y_2'= \cos x
      \end{align}

      so the Wronskian is $W = \cos^2 x + \sin^2 x = 1$. The right-hand side
      function is

      \begin{equation}
      r = \csc x
      \end{equation}

      so the particular solution is:

      \begin{align}
      y_p &=  -\cos x \int \sin x\csc(x) \d{x} + \sin x\int \cos x\csc(x) \d{x} \\
          &= -\cos x \int \d{x} + \sin x \int \frac{\cos x}{\sin x} \d{x} \\
          &= -x \cos x + \sin x \ln|x|
      \end{align}


   3. Combine and apply initial conditions:

      The general solution and its first derivative are

      \begin{align}
      y &= c_1 \cos x + c_2 \sin x \\
        & \quad + \sin x \ln\left| \sin x \right| - x \cos x \\
      y' &= -c_1 \sin x + c_2 \cos x \\
         & \quad + \cos x (1 + \ln\left| \sin x \right|) + x \sin x - \cos x
      \end{align}

      Plugging in intial conditions

      \begin{align}
      y(\pi/2) &= c_2 = 0\\
      y'(\pi/2) &= -c_1 + \pi/2 = 0
      \end{align}

      so $c_1 = \pi/2$ and $c_2 = 0$.

   The final solution is:

   \begin{equation}
   y = \frac{\pi}{2} \cos x + \sin x \ln \left| \sin x \right| - x \cos x
   \end{equation}
   ```