Variation of parameters

To solve higher-order ODEs, we’re motivated by the limitations of the method of undetermined coefficients to find a more systematic approach to find the particular solution. We use variation of parameters to do so.

The good thing with variation of parameters is that it works in principle for any forcing function input. The not so good thing is that it’s a bit more complicated as it requires integration. We may hypothetically get an unsolvable indefinite integral.

We must get our ODE in standard form:

$$y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=f(x)$$

We have no difficulty obtaining the complementary function. We assume a solution of the form, where $y_1$ and $y_2$ are the solutions to the complementary solution:

$$y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$$

After doing some tedious algebra, we get a system of equations:

$$y_1{u}_1^{\prime }+y_2{u}_2^{\prime }=0$$ $$y_1^{\prime }{u}_1^{\prime }+y_2^{\prime }{u}_2^{\prime }=f(x)$$

By Cramer’s rule, we get:

$$u_1^{\prime }=\frac{W_1}{W}=\frac{\left|\begin{array}{ccc}0& & y_2\\ f(x)& & y_2^{\prime }\end{array}\right|}{\left|\begin{array}{ccc}{y}_1& & y_2\\ y_1^{\prime }& & y_2^{\prime }\end{array}\right|}$$ $$u_2^{\prime }=\frac{W_2}{W}=\frac{\left|\begin{array}{ccc}{y}_1& & 0\\ y_1^{\prime }& & f(x)\end{array}\right|}{\left|\begin{array}{ccc}{y}_1& & y_2\\ y_1^{\prime }& & y_2^{\prime }\end{array}\right|}$$

Our resulting $u_1^{\prime }$ and $u_2^{\prime }$ can be integrated. From the assumption above, we get the particular solution. Observe that for a higher-order equation we get a much larger Wronskian determinant to evaluate. This is life, I suppose.

Addendums

As mentioned above, we may not get a function that is integrable (it is non-elementary). In this case, our function must be integral-defined (there’s not much we can do about it).