
17.7: Second Order Linear Equations II

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The method of the last section works only when the function $$f(t)$$ in $$a\ddot y+b\dot y+cy=f(t)$$ has a particularly nice form, namely, when the derivatives of $$f$$ look much like $$f$$ itself. In other cases we can try variation of parameters as we did in the first order case.

Since as before $$a\not=0$$, we can always divide by $$a$$ to make the coefficient of $$\ddot y$$ equal to 1. Thus, to simplify the discussion, we assume $$a=1$$. We know that the differential equation $$\ddot y+b\dot y+cy=0$$ has a general solution $$Ay_1+By_2$$. As before, we guess a particular solution to $$\ddot y+b\dot y+cy=f(t)$$; this time we use the guess $$y=u(t)y_1+v(t)y_2$$. Compute the derivatives:

\eqalign{ \dot y&=\dot uy_1+u\dot y_1+\dot vy_2+v\dot y_2\cr \ddot y&=\ddot uy_1+\dot u\dot y_1+\dot u\dot y_1+u\ddot y_1+\ddot vy_2+\dot v\dot y_2+\dot v\dot y_2+v\ddot y_2.\cr}

Now substituting:

\eqalign{ \ddot y+b\dot y+cy&= \ddot uy_1+\dot u\dot y_1+\dot u\dot y_1+u\ddot y_1+\ddot vy_2+\dot v\dot y_2+\dot v\dot y_2+v\ddot y_2\cr &\qquad + b\dot uy_1+bu\dot y_1+b\dot vy_2+bv\dot y_2+cuy_1+cvy_2\cr &=(u\ddot y_1+bu\dot y_1+cuy_1)+(v\ddot y_2+bv\dot y_2+cvy_2)\cr &\qquad + b(\dot uy_1+\dot vy_2) + (\ddot uy_1+\dot u\dot y_1+\ddot vy_2+\dot v\dot y_2)+ (\dot u\dot y_1+\dot v\dot y_2)\cr &=0+0+ b(\dot uy_1+\dot vy_2) + (\ddot uy_1+\dot u\dot y_1+\ddot vy_2+\dot v\dot y_2)+ (\dot u\dot y_1+\dot v\dot y_2).\cr }

The first two terms in parentheses are zero because $$y_1$$ and $$y_2$$ are solutions to the associated homogeneous equation. Now we engage in some wishful thinking. If $$\dot uy_1+\dot vy_2=0$$ then also

$\ddot uy_1+\dot u\dot y_1+\ddot vy_2+\dot v\dot y_2=0,$

by taking derivatives of both sides. This reduces the entire expression to $$\dot u\dot y_1+\dot v\dot y_2$$. We want this to be $$f(t)$$, that is, we need $$\dot u\dot y_1+\dot v\dot y_2=f(t)$$. So we would very much like these equations to be true:

\eqalign{ \dot uy_1+\dot vy_2&=0\cr \dot u\dot y_1+\dot v\dot y_2&=f(t).\cr}

This is a system of two equations in the two unknowns $$\dot u$$ and $$\dot v$$, so we can solve as usual to get $$\dot u=g(t)$$ and $$\dot v=h(t)$$.

Then we can find $$u$$ and $$v$$ by computing antiderivatives. This is of course the sticking point in the whole plan, since the antiderivatives may be impossible to find. Nevertheless, this sometimes works out and is worth a try.

Example $$\PageIndex{1}$$

Consider the equation $$\ddot y-5\dot y+6y=\sin t$$.

Solution

We can solve this by the method of undetermined coefficients, but we will use variation of parameters. The solution to the homogeneous equation is $$Ae^{2t}+Be^{3t}$$, so the simultaneous equations to be solved are

\eqalign{ \dot ue^{2t}+\dot ve^{3t}&=0\cr 2\dot ue^{2t}+3\dot ve^{3t}&=\sin t.\cr}

If we multiply the first equation by 2 and subtract it from the second equation we get

\eqalign{ \dot ve^{3t}&=\sin t\cr \dot v&=e^{-3t}\sin t\cr v&=-{1\over 10}(3\sin t+\cos t)e^{-3t},\cr}

using integration by parts. Then from the first equation:

\eqalign{ \dot u&=-e^{-2t}\dot ve^{3t}=-e^{-2t}e^{-3t}\sin(t)e^{3t}=-e^{-2t}\sin t\cr u&={1\over 5}(2\sin t+\cos t)e^{-2t}.\cr}

Now the particular solution we seek is

\eqalign{ ue^{2t}+ve^{3t}&={1\over 5}(2\sin t+\cos t)e^{-2t}e^{2t} -{1\over 10}(3\sin t+\cos t)e^{-3t}e^{3t}\cr &={1\over 5}(2\sin t+\cos t)-{1\over 10}(3\sin t+\cos t)\cr &={1\over 10}(\sin t+\cos t),\cr}

and the solution to the differential equation is

$Ae^{2t}+Be^{3t}+(\sin t+\cos t)/10.$

For comparison (and practice) you might want to solve this using the method of undetermined coefficients.

Example $$\PageIndex{2}$$:

The differential equation $$\ddot y-5\dot y+6y=e^t\sin t$$ can be solved using the method of undetermined coefficients, though we have not seen any examples of such a solution.

Solution

Again, we will solve it by variation of parameters. The equations to be solved are

\eqalign{ \dot ue^{2t}+\dot ve^{3t}&=0\cr 2\dot ue^{2t}+3\dot ve^{3t}&=e^t\sin t.\cr}

If we multiply the first equation by 2 and subtract it from the second equation we get

\eqalign{ \dot ve^{3t}&=e^t\sin t\cr \dot v&=e^{-3t}e^t\sin t=e^{-2t}\sin t\cr v&=-{1\over 5}(2\sin t+\cos t)e^{-2t}.\cr}

Then substituting we get

\eqalign{ \dot u&=-e^{-2t}\dot ve^{3t}=-e^{-2t}e^{-2t}\sin(t)e^{3t}=-e^{-t}\sin t\cr u&={1\over 2}(\sin t+\cos t)e^{-t}.\cr}

The particular solution is

\eqalign{ ue^{2t}+ve^{3t}&={1\over 2}(\sin t+\cos t)e^{-t}e^{2t} -{1\over 5}(2\sin t+\cos t)e^{-2t}e^{3t}\cr &={1\over 2}(\sin t+\cos t)e^t-{1\over 5}(2\sin t+\cos t)e^t\cr &={1\over 10}(\sin t+3\cos t)e^t,\cr}

and the solution to the differential equation is

$Ae^{2t}+Be^{3t}+e^t(\sin t+3\cos t)/10.$

Example $$\PageIndex{3}$$:

The differential equation $$\ddot y -2\dot y+y=e^t/t^2$$ is not of the form amenable to the method of undetermined coefficients. The solution to the homogeneous equation is $$Ae^t+Bte^t$$ and so the simultaneous equations are

\eqalign{ \dot ue^{t}+\dot vte^{t}&=0\cr \dot ue^{t}+\dot vte^{t}+\dot ve^t&={e^t\over t^2}.\cr}

Subtracting the equations gives

\eqalign{ \dot ve^{t}&={e^t\over t^2}\cr \dot v&={1\over t^2}\cr v&=-{1\over t}.\cr}

Then substituting we get

\eqalign{ \dot ue^t&=-\dot vte^t=-{1\over t^2}te^t\cr \dot u&=-{1\over t}\cr u&=-\ln t.\cr}

The solution is $$Ae^t+Bte^t-e^t\ln t-e^t$$.

Contributors

• Integrated by Justin Marshall.