The following method is called \(``\)Variation of Parameters."

Let \(p,q,f\) be continuous functions on some open interval \(I\). Suppose we want to find a particular solution to \(y''+py'+qy=f\) and we can find, or are given, two functions \(y_{1},y_{2}\) that form a fundamental set of solutions to \(y''+py'+qy=0\) on \(I\).

We look for a particular solution of the form \(y_p=uy_1+vy_2\) where \(u,v\) are unknown functions of the independent variable. Then \(y_p'=u'y_1+uy_1'+v'y_2+vy_2'\).

Since there are two unknown functions, we have \(``\)one degree of freedom" to use up, so we can assume that (1) \(u'y_1+v'y_2=0\) as a simplifying assumption.

So with this assumption, \(y_p'=uy_1'+vy_2'\). Then \(y_p''=u'y_1'+uy_1''+v'y_2'+vy_2''\).

Then \(y_p''+py_p'+qy_p=f\) yields \(u'y_1'+uy_1''+v'y_2'+vy_2''\,+\,p(uy_1'+vy_2')\,+\,q(uy_1+vy_2)=f\), which can be re-expressed as \(u'y_1'+v'y_2'+u(y_1''+py_1'+qy_1)+v(y_2''+py_2'+qy_2)=f\). Since \(y_1''+py_1'+qy_1=0\) and \(y_2''+py_2'+qy_2=0\) we get (2) \(u'y_1'+v'y_2'=f\).

Putting (1) and (2) together we get the following system of equations:

\begin{eqnarray*} u'y_1+v'y_2 &=& 0, \\ u'y_1'+v'y_2' &=& f. \end{eqnarray*}

Multiplying the first equation by \(y_2'\) and the second equation by \(-y_2\) yields:

\begin{eqnarray*} +u'y_1y_2'+v'y_2y_2' &=& 0, \\ -u'y_1'y_2-v'y_2y_2' &=& -fy_2. \end{eqnarray*}

Adding these equations, we get \(u'(y_1y_2'-y_1'y_2)=-fy_2\rightarrow \) \(\displaystyle u'=-\frac {fy_2}{W}\). Going back to the original system, multiplying the first equation by \(-y_1'\) and the second equation by \(y_1\) yields:

\begin{eqnarray*} -u'y_1y_1'-v'y_1'y_2 &=& 0, \\ u'y_1y_1'+v'y_1y_2' &=& fy_1. \end{eqnarray*}

Adding these equations, we get \(v'(y_1y_2'-y_1'y_2)=fy_1\rightarrow \) \(\displaystyle v'=\frac {fy_1}{W}\). Note: Always make sure the equation is in standard form as otherwise the \(``f\)" can be miss-identified.

Example: Let’s find a particular solution to \(y''+y=\tan {(t)}\) on \(I=\left (0,\frac {\pi }{2}\right )\) and then construct the general solution on \(I\).

We can use \(y_1=\cos {(t)}\) and \(y_2=\sin {(t)}\) since they form a fundamental set of solutions to the complementary equation \(y''+y=0\).

Their Wronskian is \(W=y_1y_2'-y_1'y_2=\cos ^2{(t)}+\sin ^2{(t)}=1\). So we want \(y_p=uy_1+vy_2\) where
\(\displaystyle u'=-\tan {(t)}\sin {(t)}=\frac {-\sin ^2{(t)}}{\cos {(t)}}=\frac {\cos ^2{(t)}-1}{\cos {(t)}}=\cos {(t)}-\sec {(t)}\) and
\(v'=\tan {(t)}\cos {(t)}=\sin {(t)}\).

Hence \(u=\sin {(t)}-\ln {(\tan {(t)}+\sec {(t)})}\) and \(v=-\cos {(t)}\).

So \(y_p=uy_1+vy_2=(\sin {(t)}-\ln {(\tan {(t)}+\sec {(t)})})\cos {(t)}-\cos {(t)}\sin {(t)}=-\cos {(t)}\ln {(\tan {(t)}+\sec {(t)})})\) is particular solution to \(y''+y=\tan {(t)}\), and the general solution is
\(y=-\cos {(t)}\ln {(\tan {(t)}+\sec {(t)})})+c_1\cos {(t)}+c_2\sin {(t)}\).

Another Example: Let’s find a particular solution to \(x^2 y''-2xy'+2y=5x\) on \(I=(0,\infty )\) and then construct the general solution on \(I\).

The Euler-characteristic polynomial of \(x^2 y''-2xy'+2y=0\) is \(m(m-1)-2m+2\), equiv. \(m^2-3m+2=(m-1)(m-2)\). Since it has roots at \(m=1\) and \(m=2\) we can use \(y_1=x\) and \(y_2=x^2\) as a fundamental set of solutions to the complementary equation on \(I\).

Their Wronskian is \(W=y_1y_2'-y_1'y_2=x(2x)-x^2=x^2\). We carefully determine that \(\displaystyle f=\frac {5}{x}\). Then \(\displaystyle u'=-\frac {(5/x)x^2}{x^2}=-\frac {5}{x}\) and \(\displaystyle v'=\frac {(5/x)x}{x^2}=\frac {5}{x^2}\). Then \(u=-5\ln {(x)}\) and \(\displaystyle v=-\frac {5}{x}\). So \(\displaystyle y_p=uy_1+vy_2=-5x\ln {(x)}-5x\). Hence the general solution is \(y=-5x\ln {(x)}-5x+c_1x+c_2x^2\). Or equivalently
\(y=-5x\ln {(x)}+c_1x+c_2x^2\).