6. Variation of parameters

The preceding section explained how to find the general solution \(x_h(t)\) of a homoegenous linear system \(\bfA\bfx=\bfzero\). The next step of our overall solution strategy is to find any particular solution of \(\opA[x]=f\) for a given \(f(t)\).

The form of the homogeneous solution is

\[ x_h(t) = c_1 g(t), \]

where \(c_1\) is an arbitrary constant and \(g(t)\) is computed as the exponentiation of an integral. Let’s see what happens if we replace the constant by an unknown function of \(t\), \(x_p(t)=k(t)g(t)\):

\[\begin{align*} \opA[x_p] &= x_p' - a(t) x_p \\ & = (k'g+kg') - a k g \\ & = k'g + k(g'-ag) \\ &= k'g + k\cdot\opA[g]. \end{align*}\]

However, \(g\) is itself a homogeneous solution, so all that remains is \(k'g\). We therefore make \(x_p\) a particular solution if we set \(f=k'g\). This lets us solve for the unknown \(k(t)\) by taking \(k'(t)=f(t)/g(t)\) and integrating. The result is known as the variation of parameters formula (or VoP formula for short). We collect all the equations here, including the homogeneous and general solutions.

Formula 6.2 (Variation of parameters)

To find a particular solution \(x_p\) of \(\opA[x]=f\), compute

\[\begin{align*} g(t) &= \exp\left[ \int a(t)\, dt\right], \\ k(t) &= \int \frac{f(t)}{g(t)}\, dt, \\ x_p(t) &= k(t)g(t), \\ x(t) &= c_1g(t) + x_p(t). \end{align*}\]

Note

The integration constants you would normally get from the indefinite integrals in the VoP formula can be ignored (i.e., set to zero), as they make no meaningful difference in the final result. They are “absorbed” into \(c_1\).

Example

Solve \(3x'=12x+24t\).

Solution

Rewriting the ODE as \(x'-4x=8t\), we identify \(a(t)=4\) and \(f(t)=8t\). Then

\[g(t) = \exp\left[ \int 4\,dt \right] = e^{4t},\]

and

\[k(t) = \int \frac{8t}{e^{4t}}\, dt = -\frac{1}{2} (4t+1)e^{-4t},\]

where you need integration by parts (or a computer) to perform the integral. Hence

\[x_p(t) = \left[ \frac{1}{2} (4t+1)e^{-4t} \right] e^{4t} = -\frac{1}{2} (4t+1),\]

and the general solution is

\[x(t) = x_h(t) + x_p(t) = c_1 e^{4t} - \frac{1}{2} (4t+1).\]

Example

Find a particular solution of \(x'=2t x + 6t\).

Solution

We start with

\[g(t) = \exp\left[ \int 2t\, dt \right] = e^{t^2}.\]

Then we can compute

\[k(t) = \int \frac{6t}{e^{t^2}}\, dt = -3 e^{-t^2}.\]

Finally,

\[x_p(t) = k(t) g(t) = -3,\]

and the general solution is \(x(t)=c_1 e^{t^2}-3\).

Example

Solve the IVP

\[ (2+t) x'= x - 1, \quad x(0) = -5. \]
Solution

First we put the ODE into our standard form,

\[x' - \frac{1}{2+t} x = -\frac{1}{2+t}.\]

Then

\[g(t) = \exp\left[ \int \frac{1}{2+t}\, dt \right] = \exp[ \ln(2+t) ] = 2+t.\]

Next,

\[k(t) = \int \frac{-1}{2+t} (2+t)^{-1} \, dt = (2+t)^{-1},\]

so that the general solution is

\[x(t) = c_1 (2+t) + (2+t)^{-1} (2+t) = c_1(2+t)+1.\]

Finally, we apply the initial condition to solve for \(c_1\):

\[ -5 = x(0) = 2c_1+1 \quad \implies \quad c_1=-3. \]

Hence \(x(t) = 1-3(2+t) = -5-3t.\)

Attention

It takes a pretty special relationship between \(a(t)\) and \(f(t)\) to make the integrals in the VoP formula reasonable to do by hand. Consequently, if you are asked in an exercise to produce a solution and make a mistake, you will likely wind up with an integral that is virually impossible to evaluate.

5. Homogeneous solutions 7. IVP formula