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Mathematics LibreTexts

17.7: Second Order Linear Equations II

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

David Guichard (Whitman College)

  • Integrated by Justin Marshall.