Use method of variations of parameters to find particular solution and check your solution using method of undetermined coefficients. \(y^{\prime \prime }-5y^{\prime }+6y=2e^{t}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }-5y^{\prime }+6y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters and also using undetermined coefficients to compare with.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}-5r+6=0\) or \(\left ( r-3\right ) \left ( r-2\right ) =0.\) Therefore the roots are \(r_{1}=3,r_{2}=2\). Hence the two fundamental solutions are \begin{align*} y_{1} & =e^{3t}\\ y_{2} & =e^{2t} \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{3t}+c_{2}e^{2t} \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\[ W\left ( t\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{3t} & e^{2t}\\ 3e^{3t} & 2e^{2t}\end{vmatrix} =2e^{5t}-3e^{5t}=-e^{5t}\] Letting \(g\left ( t\right ) =2e^{t}\) therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W}dt=-\int \frac{e^{2t}2e^{t}}{-e^{5t}}dt=2\int \frac{e^{3t}}{e^{5t}}dt=2\int e^{-2t}dt=2\left [ \frac{e^{-2t}}{-2}\right ] =-e^{-2t}\] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W}dt=\int \frac{e^{3t}2e^{t}}{-e^{5t}}dt=-2\int \frac{e^{4t}}{e^{5t}}dt=-2\int e^{-t}dt=-2\left [ \frac{e^{-t}}{-1}\right ] =2e^{-t}\] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( -e^{-2t}\right ) e^{3t}+2e^{-t}e^{2t}\\ & =-e^{t}+2e^{t}\\ & =e^{t} \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{3t}+c_{2}e^{2t}+e^{t} \end{align*}
Finding \(y_{p}\) using undetermined coefficients
From the form of \(g\left ( t\right ) \) in the problem, particular solution is assumed to be\[ y_{p}=Ae^{t}\] Hence\begin{align*} y_{p}^{\prime } & =Ae^{t}\\ y_{p}^{\prime \prime } & =Ae^{t} \end{align*}
Plugging back into the original ODE gives\begin{align*} y_{p}^{\prime \prime }-5y_{p}^{\prime }+6y_{p} & =2e^{t}\\ Ae^{t}-5Ae^{t}+6Ae^{t} & =2e^{t} \end{align*}
Dividing by \(e^{t}\neq 0\) gives\begin{align*} A-5A+6A & =2\\ 2A & =2\\ A & =1 \end{align*}
Therefore\[ y_{p}=e^{t}\] Which agrees with variation of parameters particular solution found earlier. Therefore the same general solution is obtained as expected. QED.
Use method of variations of parameters to find particular solution and check your solution using method of undetermined coefficients. \(y^{\prime \prime }-y^{\prime }-2y=2e^{-t}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }-y^{\prime }-2y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters and also using undetermined coefficients to compare with.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}-r-2=0\) or \(\left ( r+1\right ) \left ( r-2\right ) =0.\) Therefore the roots are \(r_{1}=-1,r_{2}=2\). Hence the two fundamental solutions are \begin{align*} y_{1} & =e^{-t}\\ y_{2} & =e^{2t} \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{-t}+c_{2}e^{2t} \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\[ W\left ( t\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{-t} & e^{2t}\\ -e^{-t} & 2e^{2t}\end{vmatrix} =2e^{t}+e^{t}=3e^{t}\] Letting \(g\left ( t\right ) =2e^{-t}\) therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W}dt=-\int \frac{e^{2t}2e^{-t}}{3e^{t}}dt=-\frac{2}{3}\int \frac{e^{t}}{e^{t}}dt=-\frac{2}{3}t \] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W}dt=\int \frac{e^{-t}2e^{-t}}{3e^{t}}dt=\frac{2}{3}\int \frac{e^{-2t}}{e^{t}}dt=\frac{2}{3}\int e^{-3t}dt=\frac{2}{3}\left [ \frac{e^{-3t}}{-3}\right ] =-\frac{2}{9}e^{-3t}\] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( -\frac{2}{3}t\right ) e^{-t}-\frac{2}{9}e^{-3t}e^{2t}\\ & =-\frac{2}{3}te^{-t}-\frac{2}{9}e^{-t} \end{align*}
We notice something here. The extra term \(-\frac{2}{9}e^{-t}\) above is constant times one of the fundamental solutions (one of the solutions to the homogenous equation), which is \(y_{1}\) in this case found earlier. But adding a multiple of a fundamental solution to a particular solution gives another particular solution. So the term \(-\frac{2}{9}e^{-t}\) will be merged with the term from the homogenous solution. Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-t}+c_{2}e^{2t}-\frac{2}{3}te^{-t}-\frac{2}{9}e^{-t} \end{align*}
We can now combine \(\frac{2}{9}e^{-t}\) that shows up from the particular solution with the \(c_{1}e^{-t}\) term from the homogenous solution, since \(c_{1}\) is arbitrary constant, which simplifies the above to\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-t}+c_{2}e^{2t}-\frac{2}{3}te^{-t} \end{align*}
Finding \(y_{p}\) using undetermined coefficients
From the form of \(g\left ( t\right ) \) in the problem, and since \(e^{-t}\) is already one of the fundamental solutions, then particular solution is assumed to be\[ y_{p}=Ate^{-t}\] Hence\begin{align*} y_{p}^{\prime } & =A\left ( e^{-t}-te^{-t}\right ) \\ y_{p}^{\prime \prime } & =A\left ( -e^{-t}-e^{-t}+te^{-t}\right ) \\ & =A\left ( -2e^{-t}+te^{-t}\right ) \end{align*}
Plugging back into the original ODE gives\begin{align*} y_{p}^{\prime \prime }-y_{p}^{\prime }-2y_{p} & =2e^{-t}\\ A\left ( -2e^{-t}+te^{-t}\right ) -A\left ( e^{-t}-te^{-t}\right ) -2Ate^{-t} & =2e^{-t} \end{align*}
Dividing by \(e^{-t}\neq 0\) gives\begin{align*} A\left ( -2+t\right ) -A\left ( 1-t\right ) -2At & =2\\ t\left ( A+A-2A\right ) -2A-A & =2\\ -3A & =2\\ A & =\frac{-2}{3} \end{align*}
Therefore\[ y_{p}=\frac{-2}{3}te^{-t}\] Which agrees with variation of parameters particular solution found earlier. Therefore the same general solution is obtained as expected. QED.
Use method of variations of parameters to find particular solution and check your solution using method of undetermined coefficients. \(y^{\prime \prime }+2y^{\prime }+y=3e^{-t}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }+2y^{\prime }+y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters and also using undetermined coefficients to compare with.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}+2r+1=0\) or \(\left ( r+1\right ) \left ( r+1\right ) =0\), Therefore the roots are duplicate \(r_{1}=-1\). Hence the two fundamental solutions are \begin{align*} y_{1} & =e^{-t}\\ y_{2} & =te^{-t} \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{-t}+c_{2}te^{-t} \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\begin{align*} W\left ( t\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{-t} & te^{-t}\\ -e^{-t} & e^{-t}-te^{-t}\end{vmatrix} \\ & =\left ( e^{-t}\right ) \left ( e^{-t}-te^{-t}\right ) +\left ( te^{-t}\right ) \left ( e^{-t}\right ) \\ & =e^{-2t}-te^{-2t}+te^{-2t}\\ & =e^{-2t} \end{align*}
Letting \(g\left ( t\right ) =3e^{-t}\) therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W}dt=-\int \frac{te^{-t}\left ( 3e^{-t}\right ) }{e^{-2t}}dt=-3\int tdt=-\frac{3}{2}t^{2}\] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W}dt=\int \frac{e^{-t}\left ( 3e^{-t}\right ) }{e^{-2t}}dt=3\int dt=3t \] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( -\frac{3}{2}t^{2}\right ) e^{-t}+3t\left ( te^{-t}\right ) \\ & =-\frac{3}{2}t^{2}e^{-t}+3t^{2}e^{-t}\\ & =\frac{3}{2}t^{2}e^{-t} \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-t}+c_{2}te^{-t}+\frac{3}{2}t^{2}e^{-t} \end{align*}
Finding \(y_{p}\) using undetermined coefficients
From the form of \(g\left ( t\right ) =3e^{-t}\) in the problem, we want to try \(e^{-t}\) but since \(e^{-t}\) is already one of the fundamental solutions, we then look at \(te^{-t}\) but this is also one fundamental solutions, then we look for \(t^{2}e^{-t}\). Hence\[ y_{p}=At^{2}e^{-t}\] Hence\begin{align*} y_{p}^{\prime } & =A\left ( 2te^{-t}-t^{2}e^{-t}\right ) \\ y_{p}^{\prime \prime } & =A\left ( 2e^{-t}-2te^{-t}-\left ( 2te^{-t}-t^{2}e^{-t}\right ) \right ) \\ & =A\left ( 2e^{-t}-2te^{-t}-2te^{-t}+t^{2}e^{-t}\right ) \\ & =A\left ( 2e^{-t}-4te^{-t}+t^{2}e^{-t}\right ) \end{align*}
Plugging back into the original ODE gives\begin{align*} y_{p}^{\prime \prime }+2y_{p}^{\prime }+y_{p} & =3e^{-t}\\ A\left ( 2e^{-t}-4te^{-t}+t^{2}e^{-t}\right ) +2A\left ( 2te^{-t}-t^{2}e^{-t}\right ) +At^{2}e^{-t} & =3e^{-t} \end{align*}
Dividing by \(e^{-t}\neq 0\) gives\begin{align*} A\left ( 2-4t+t^{2}\right ) +2A\left ( 2t-t^{2}\right ) +At^{2} & =3\\ t\left ( -4A+4A\right ) +t^{2}\left ( A-2A+A\right ) +2A & =3\\ A & =\frac{3}{2} \end{align*}
Therefore\[ y_{p}=\frac{3}{2}te^{-t}\] Which agrees with variation of parameters particular solution found earlier. Therefore the same general solution is obtained as expected. QED.
Use method of variations of parameters to find particular solution and check your solution using method of undetermined coefficients. \(4y^{\prime \prime }-4y^{\prime }+y=16e^{\frac{t}{2}}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(4y^{\prime \prime }-4y^{\prime }+y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters and also using undetermined coefficients to compare with.
Finding \(y_{h}\)
The first step is to put the ODE in standard form, with the coefficient of \(y^{\prime \prime }\) being one. Hence it becomes\[ y^{\prime \prime }-y^{\prime }+\frac{1}{4}y=4e^{\frac{t}{2}}\] Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}-r+\frac{1}{4}=0\) or \(\left ( r-\frac{1}{2}\right ) \left ( r-\frac{1}{2}\right ) =0\), Therefore the roots are duplicate \(r=\frac{1}{2}\). Hence the two fundamental solutions are \begin{align*} y_{1} & =e^{\frac{1}{2}t}\\ y_{2} & =te^{\frac{1}{2}t} \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{\frac{1}{2}t}+c_{2}te^{\frac{1}{2}t} \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\begin{align*} W\left ( t\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{\frac{1}{2}t} & te^{\frac{1}{2}t}\\ \frac{1}{2}e^{\frac{1}{2}t} & e^{\frac{1}{2}t}+\frac{1}{2}te^{\frac{1}{2}t}\end{vmatrix} \\ & =\left ( e^{\frac{1}{2}t}\right ) \left ( e^{\frac{1}{2}t}+\frac{1}{2}te^{\frac{1}{2}t}\right ) -\left ( te^{\frac{1}{2}t}\right ) \left ( \frac{1}{2}e^{\frac{1}{2}t}\right ) \\ & =e^{t}+\frac{1}{2}te^{t}-\frac{1}{2}te^{t}\\ & =e^{t} \end{align*}
Letting \(g\left ( t\right ) =4e^{\frac{t}{2}}\) therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W}dt=-\int \frac{te^{\frac{1}{2}t}\left ( 4e^{\frac{t}{2}}\right ) }{e^{t}}dt=-4\int tdt=-2t^{2}\] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W}dt=\int \frac{e^{\frac{1}{2}t}\left ( 4e^{\frac{t}{2}}\right ) }{e^{t}}dt=4\int dt=4t \] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( -2t^{2}\right ) e^{\frac{1}{2}t}+4t\left ( te^{\frac{1}{2}t}\right ) \\ & =-2t^{2}e^{\frac{1}{2}t}+4t^{2}e^{\frac{1}{2}t}\\ & =2t^{2}e^{\frac{1}{2}t} \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{\frac{1}{2}t}+c_{2}te^{\frac{1}{2}t}+2t^{2}e^{\frac{1}{2}t} \end{align*}
Finding \(y_{p}\) using undetermined coefficients
From the form of \(g\left ( t\right ) =4e^{\frac{t}{2}}\) in the problem, we want to try \(e^{\frac{t}{2}}\) but since \(e^{\frac{t}{2}}\) is already one of the fundamental solutions, we then look at \(te^{\frac{t}{2}}\) but this is also one fundamental solutions, then we look for \(t^{2}e^{\frac{t}{2}}\). Hence\[ y_{p}=At^{2}e^{\frac{t}{2}}\] Hence\begin{align*} y_{p}^{\prime } & =A\left ( 2te^{\frac{t}{2}}+\frac{1}{2}t^{2}e^{\frac{t}{2}}\right ) \\ y_{p}^{\prime \prime } & =A\left ( 2e^{\frac{t}{2}}+te^{\frac{t}{2}}+te^{\frac{t}{2}}+\frac{1}{4}t^{2}e^{\frac{t}{2}}\right ) \\ & =A\left ( 2e^{\frac{t}{2}}+2te^{\frac{t}{2}}+\frac{1}{4}t^{2}e^{\frac{t}{2}}\right ) \end{align*}
Plugging back into the original ODE gives\begin{align*} y_{p}^{\prime \prime }-y_{p}^{\prime }+\frac{1}{4}y_{p} & =4e^{\frac{t}{2}}\\ A\left ( 2e^{\frac{t}{2}}+2te^{\frac{t}{2}}+\frac{1}{4}t^{2}e^{\frac{t}{2}}\right ) -A\left ( 2te^{\frac{t}{2}}+\frac{1}{2}t^{2}e^{\frac{t}{2}}\right ) +\frac{1}{4}At^{2}e^{\frac{t}{2}} & =4e^{\frac{t}{2}} \end{align*}
Dividing by \(e^{\frac{t}{2}}\neq 0\) gives\begin{align*} A\left ( 2+2t+\frac{1}{4}t^{2}\right ) -A\left ( 2t+\frac{1}{2}t^{2}\right ) +\frac{1}{4}At^{2} & =4\\ t\left ( 2A-2A\right ) +t^{2}\left ( \frac{1}{4}A-\frac{1}{2}A+\frac{1}{4}A\right ) +2A & =4\\ A & =2 \end{align*}
Therefore\[ y_{p}=2t^{2}e^{\frac{t}{2}}\] Which agrees with variation of parameters particular solution found earlier. Therefore the same general solution is obtained as expected. QED.
Find the general solution of \(y^{\prime \prime }+y=\tan t\) for \(0<t<\frac{\pi }{2}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }+y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}+1=0\) or \(r=\pm i\). Hence the two fundamental solutions are \begin{align*} y_{1} & =\cos t\\ y_{2} & =\sin t \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\cos t+c_{2}\sin t \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\[ W\left ( t\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} \cos t & \sin t\\ -\sin t & \cos t \end{vmatrix} =\cos ^{2}t+\sin ^{2}t=1 \] Let \(g\left ( t\right ) =\tan t\), therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\begin{align*} u_{1}\left ( t\right ) & =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=-\int \frac{\sin t\tan t}{1}dt=-\int \sin t\frac{\sin t}{\cos t}dt=-\int \frac{\sin ^{2}t}{\cos t}dt\\ & =-\int \frac{1-\cos ^{2}t}{\cos t}dt=\int \frac{\cos ^{2}t-1}{\cos t}dt=\int \cos t-\frac{1}{\cos t}dt\\ & =\int \cos tdt-\int \frac{1}{\cos t}dt\\ & =\sin t-\int \sec tdt\\ & =\sin t-\ln \left ( \sec (t)+\tan (t)\right ) \end{align*}
And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=\int \frac{\cos t\tan t}{1}dt=\int \cos t\frac{\sin t}{\cos t}\ dt=\int \sin t\ dt=-\cos t \] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( \sin t-\ln \left ( \sec (t)+\tan (t)\right ) \right ) \cos t+\left ( -\cos t\right ) \sin t\\ & =-\cos \left ( t\right ) \ln \left ( \sec (t)+\tan (t)\right ) \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\cos t+c_{2}\sin t-\cos \left ( t\right ) \ln \left ( \sec (t)+\tan (t)\right ) \end{align*}
Find the general solution of \(y^{\prime \prime }+9y=9\sec ^{2}3t\) for \(0<t<\frac{\pi }{6}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }+9y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}+9=0\) or \(r=\pm 3i\). Hence the two fundamental solutions are \begin{align*} y_{1} & =\cos 3t\\ y_{2} & =\sin 3t \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\cos 3t+c_{2}\sin 3t \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\[ W\left ( t\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} \cos 3t & \sin 3t\\ -3\sin 3t & 3\cos 3t \end{vmatrix} =3\cos ^{2}t+3\sin ^{2}t=3 \] Let \(g\left ( t\right ) =\frac{9}{\cos ^{2}3t}\), therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=-\int \frac{9\sin \left ( 3t\right ) }{3\cos ^{2}\left ( 3t\right ) }dt=-3\int \frac{\sin \left ( 3t\right ) }{\cos ^{2}\left ( 3t\right ) }dt \] Let \(u=\cos \left ( 3t\right ) \), hence \(\frac{du}{dt}=-3\sin 3t\rightarrow dt=\frac{du}{-3\sin 3t}\) and the above integral becomes\[ u_{1}\left ( t\right ) =-3\int \frac{\sin \left ( 3t\right ) }{u^{2}}\frac{du}{-3\sin 3t}=\int \frac{1}{u^{2}}du=\frac{-1}{u}=\frac{-1}{\cos 3t}=-\sec \left ( 3t\right ) \] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=\int \frac{9\cos 3t}{3\cos ^{2}\left ( 3t\right ) }dt=3\int \frac{1}{\cos \left ( 3t\right ) }\ dt=3\int \sec \left ( 3t\right ) \ dt=\ln \left ( \sec (3t)+\tan (3t)\right ) \] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =-\sec \left ( 3t\right ) \cos 3t+\ln \left ( \sec (3t)+\tan (3t)\right ) \sin 3t\\ & =-1+\ln \left ( \sec (t)+\tan (t)\right ) \sin 3t \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\cos 3t+c_{2}\sin 3t-1+\sin 3t\ln \left ( \sec (t)+\tan (t)\right ) \end{align*}
Find the general solution of \(y^{\prime \prime }+4y^{\prime }+4y=t^{-2}e^{-2t}\) for \(t>0\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }+4y^{\prime }+4y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}+4r+4=0\) or \(\left ( r+2\right ) \left ( r+2\right ) =0\). Hence double root \(r=-2\) and the fundamental solutions are \begin{align*} y_{1} & =e^{-2t}\\ y_{2} & =te^{-2t} \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{-2t}+c_{2}te^{-2t} \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\begin{align*} W\left ( t\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{-2t} & te^{-2t}\\ -2e^{-2t} & e^{-2t}-2te^{-2t}\end{vmatrix} =e^{-2t}\left ( e^{-2t}-2te^{-2t}\right ) +2e^{-2t}\left ( te^{-2t}\right ) \\ & =e^{-4t}-2te^{-4t}+2te^{-4t}\\ & =e^{-4t} \end{align*}
Let \(g\left ( t\right ) =t^{-2}e^{-2t}\), therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=-\int \frac{te^{-2t}t^{-2}e^{-2t}}{e^{-4t}}dt=-\int t^{-1}dt=-\ln \left \vert t\right \vert \] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=\int \frac{e^{-2t}t^{-2}e^{-2t}}{e^{-4t}}dt=\int t^{-2}dt=-\frac{1}{t}\] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =-\ln \left \vert t\right \vert e^{-2t}-\frac{1}{t}te^{-2t}\\ & =-e^{-2t}\ln \left \vert t\right \vert -e^{-2t} \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-2t}+c_{2}te^{-2t}-e^{-2t}\ln \left \vert t\right \vert -e^{-2t} \end{align*}
We can combine \(e^{-2t}\) that shows up from the particular solution with the \(c_{1}e^{-2t}\) term from the homogenous solution, since \(c_{1}\) is arbitrary constant, which simplifies the above to\[ y=c_{1}e^{-2t}+c_{2}te^{-2t}-e^{-2t}\ln \left \vert t\right \vert \]
Find the general solution of \(y^{\prime \prime }+4y=3\frac{1}{\sin 2t}\) for \(0<t<\frac{\pi }{2}\)
solution
The general solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }+4y=0\) and \(y_{p}\) is a particular solution which is found using variations of parameters.
Finding \(y_{h}\)
Since ODE has constant coefficients, then the characteristic equation is used. It is given by \(r^{2}+4=0\) or \(r=\pm 2i\). The fundamental solutions are \begin{align*} y_{1} & =\cos 2t\\ y_{2} & =\sin 2t \end{align*}
And the homogenous solution is therefore given by\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\cos 2t+c_{2}\sin 2t \end{align*}
Finding \(y_{p}\) using variation of parameters
First step is to find Wronskian \(W\) given by\[ W\left ( t\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} \cos 2t & \sin 2t\\ -2\sin 2t & 2\cos 2t \end{vmatrix} =2\cos ^{2}2t+2\sin ^{2}2t=2 \] Let \(g\left ( t\right ) =\frac{3}{\sin 2t}\), therefore the particular solution is\[ y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) \] Where\[ u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=-\int \frac{\sin \left ( 2t\right ) 3}{2\sin 2t}dt=-\frac{3}{2}\int dt=\frac{-3}{2}t \] And\[ u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) g\left ( t\right ) }{W\left ( t\right ) }dt=\int \frac{\cos \left ( 2t\right ) 3}{2\sin 2t}dt=\frac{3}{2}\int \frac{\cos \left ( 2t\right ) }{\sin \left ( 2t\right ) }dt \] Let \(u=\sin 2t\rightarrow du=2\cos 2tdt\) and the above integral becomes\[ u_{2}\left ( t\right ) =\frac{3}{2}\int \frac{\cos \left ( 2t\right ) }{u}\frac{du}{2\cos 2t}=\frac{3}{4}\int \frac{1}{u}du=\frac{3}{4}\ln \left \vert u\right \vert =\frac{3}{4}\ln \left \vert \sin 2t\right \vert \] Hence the particular solution becomes\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\frac{-3}{2}t\cos 2t+\frac{3}{4}\ln \left \vert \sin 2t\right \vert \sin 2t \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\cos 2t+c_{2}\sin 2t-\frac{3}{2}t\cos 2t+\frac{3}{4}\sin \left ( 2t\right ) \ln \left \vert \sin 2t\right \vert \end{align*}