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5. Homogeneous solutions

Step one of our overall solution strategy is to solve the homogeneous problem \(\opA[x]=0\), or

\[ \dd{x}{t} = a(t) x. \]

Let’s riff on the exponential solution we found with constant \(a\) by guessing that

\[ x(t) = \exp\bigl[q(t)\bigr], \]

for some \(q(t)\) yet to be determined. Note that

\[ x'(t) = q'(t) e^{\,q(t)} = q'(t) x. \]

If this is to be a solution of \(x'=ax\), all we need is that \(q'(t)=a(t)\). This is solved by simple integration. To summarize:

Formula 5.4 (Solution of \(x'=a(t)x\))
\[\begin{split} \begin{align*} q(t) &= \int a(t) \, dt, \\ x(t) &= \exp\bigl[q(t)\bigr]. \end{align*} \end{split}\]

Example

Solve \(x'=\sin(t) x\).

Solution

We integrate \(a(t)=\sin(t)\) to get

\[ q(t) = -\cos(t)+C. \]

The general solution is thus \(e^{\,q(t)}\), or

\[x_h(t) = e^C \exp[ -\cos(t) ] = c_1 \exp[ -\cos(t) ].\]

Note

As seen in the preceding example, the integration constant pops up additively in \(q\) and then multiplicatively in \(x\). You can ignore the integration constant until the end and then just tack it on; remember that by linearity, any constant multiple of a homogeneous solution is also a solution.

Example

Solve the homogeneous IVP

\[ t x' = 2x, \quad x(2) = 12. \]
Solution

First we rewrite the ODE as

\[ x' = \frac{2x}{t}, \]

from which we see that \(a(t)=2/t\). We first find the general solution by integration:

\[ x_h(t) = c_1 \exp\left[ \int 2t^{-1}\, dt \right] = c_1 \exp\bigl[2\ln(t)\bigr] = c_1 t^2. \]

This is the general solution. To eliminate the integration constant, we apply the initial condition:

\[ 12 = x_h(2) = c_1\cdot 2^2. \]

Hence \(x(t) = 3t^2\).

5.1. Initial-value formula

We can take a shortcut in the case of a homogeneous IVP that avoids introducing an integration constant and then eliminating it. The trick is to use a definite integral to compute \(q\), as in

\[ \tilde{q}(t) = \int_{t_0}^t a(s)\, ds. \]

The advantage of doing so is that by its construction, \(\tilde{q}(t_0)=0\). Therefore if we define

\[ x_h(t)= c_1\exp\bigl[\tilde{q}(t)\bigr], \]

we find that \(x(t_0)=c_1\exp(0)=c_1\).

Formula 5.5 (Solution of \(x' = a(t)x,\) \(x(t_0)=x_0\))
\[ x(t) = x_0 \exp\left[ \int_{t_0}^t a(s)\, ds \right]. \]

Example (continued)

We revisit the preceding example of

\[ x' = \frac{2x}{t}, \quad x(2) = 12, \]

and apply the new formula to solve it:

\[\begin{align*} x(t) &= 12 \exp\left[ \int_2^t 2s^{-1}\, ds \right] \\ &= 12 \exp\bigl[ 2\ln(t) - 2\ln(2) \bigr] \\ &= \frac{12t^2}{4}. \end{align*}\]

This is the same result as before.

4. Operators 6. Variation of parameters