Problem Make the substitution \(v=\ln x\) to find general solution for \(x>0\) of the Euler equation \(x^{2}y^{\prime \prime }+xy^{\prime }-y=0\)
solution Let \(v=\ln x\). Hence \(\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx}=\frac{dy}{dv}\frac{1}{x}\) and \begin{align*} \frac{d^{2}y}{dx^{2}} & =\frac{d}{dx}\left ( \frac{dy}{dv}\frac{1}{x}\right ) \\ & =\frac{d^{2}y}{dv^{2}}\frac{dv}{dx}\frac{1}{x}+\frac{dy}{dv}\frac{d}{dx}\left ( \frac{1}{x}\right ) \\ & =\frac{d^{2}y}{dv^{2}}\frac{1}{x^{2}}-\frac{dy}{dv}\frac{1}{x^{2}} \end{align*}
Hence the ODE becomes\begin{align*} x^{2}y^{\prime \prime }+xy^{\prime }-y & =0\\ x^{2}\left ( \frac{d^{2}y}{dv^{2}}\frac{1}{x^{2}}-\frac{dy}{dv}\frac{1}{x^{2}}\right ) +x\left ( \frac{dy}{dv}\frac{1}{x}\right ) -y\left ( v\right ) & =0\\ \frac{d^{2}y}{dv^{2}}-\frac{dy}{dv}+\frac{dy}{dv}-y\left ( v\right ) & =0\\ \frac{d^{2}y}{dv^{2}}-y\left ( v\right ) & =0 \end{align*}
This can now be solved using characteristic equation. \(r^{2}-1=0\) or \(r^{2}=1\) or \(r=\pm 1\). Hence the solution is\[ y\left ( v\right ) =c_{1}e^{v}+c_{2}e^{-v}\] But \(v=\ln x\), hence\begin{align*} y\left ( x\right ) & =c_{1}e^{\ln x}+c_{2}e^{-\ln x}\\ & =c_{1}x+c_{1}\frac{1}{x} \end{align*}
The above is the solution.
But an easier method is the following. Let \(y=x^{r}\). Hence \(y^{\prime }=rx^{r-1},y^{\prime \prime }=r\left ( r-1\right ) x^{r-2}\). Substituting this into the ODE gives\begin{align*} r\left ( r-1\right ) x^{r}+rx^{r}-x^{r} & =0\\ x^{r}\left ( r\left ( r-1\right ) +r-1\right ) & =0 \end{align*}
Since \(x^{r}\neq 0\), we simplify the above and obtain the characteristic equation\begin{align*} r\left ( r-1\right ) +r-1 & =0\\ r^{2}-1 & =0\\ r^{2} & =1\\ r & =\pm 1 \end{align*}
Hence \begin{align*} y\left ( x\right ) & =c_{1}x^{r_{1}}+c_{2}x^{r_{2}}\\ & =c_{1}x+c_{2}x^{-1} \end{align*}
For \(x>0\).
Problem Make the substitution \(v=\ln x\) to find general solution for \(x>0\) of the Euler equation \(4x^{2}y^{\prime \prime }+8xy^{\prime }-3y=0\)
solution Let \(v=\ln x\). Hence \(\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx}=\frac{dy}{dv}\frac{1}{x}\) and \begin{align*} \frac{d^{2}y}{dx^{2}} & =\frac{d}{dx}\left ( \frac{dy}{dv}\frac{1}{x}\right ) \\ & =\frac{d^{2}y}{dv^{2}}\frac{dv}{dx}\frac{1}{x}+\frac{dy}{dv}\frac{d}{dx}\left ( \frac{1}{x}\right ) \\ & =\frac{d^{2}y}{dv^{2}}\frac{1}{x^{2}}-\frac{dy}{dv}\frac{1}{x^{2}} \end{align*}
Hence the ODE becomes\begin{align*} x^{2}y^{\prime \prime }+xy^{\prime }-y & =0\\ 4x^{2}\left ( \frac{d^{2}y}{dv^{2}}\frac{1}{x^{2}}-\frac{dy}{dv}\frac{1}{x^{2}}\right ) +8x\left ( \frac{dy}{dv}\frac{1}{x}\right ) -3y\left ( v\right ) & =0\\ 4\frac{d^{2}y}{dv^{2}}-4\frac{dy}{dv}+8\frac{dy}{dv}-3y\left ( v\right ) & =0\\ 4\frac{d^{2}y}{dv^{2}}+4\frac{dy}{dv}-3y\left ( v\right ) & =0 \end{align*}
This can now be solved using characteristic equation. \(4r^{2}+4r-3=0\), whose roots are \(r_{1}=\frac{-3}{2},r_{2}=\frac{1}{2}\) Hence the solution is\[ y\left ( v\right ) =c_{1}e^{\frac{-3}{2}v}+c_{2}e^{\frac{1}{2}v}\] But \(v=\ln x\), hence\begin{align*} y\left ( x\right ) & =c_{1}e^{\frac{-3}{2}\ln x}+c_{2}e^{\frac{1}{2}\ln x}\\ & =c_{1}x^{\frac{-3}{2}}+c_{1}x^{\frac{1}{2}} \end{align*}
Problem Use reduction of order to find second L.I. solution \(y_{2}\). \(x^{2}y^{\prime \prime }-x\left ( x+2\right ) y^{\prime }+\left ( x+2\right ) y=0\) with \(y_{1}=x\) and \(x>0\)
solution Let \(y=vy_{1}\), hence\begin{align*} y^{\prime } & =v^{\prime }y_{1}+vy_{1}^{\prime }\\ y^{\prime \prime } & =v^{\prime \prime }y_{1}+v^{\prime }y_{1}^{\prime }+v^{\prime }y_{1}^{\prime }+vy_{1}^{\prime \prime }\\ & =v^{\prime \prime }y_{1}+2v^{\prime }y_{1}^{\prime }+vy_{1}^{\prime \prime } \end{align*}
Therefore the original ODE becomes\begin{align*} x^{2}y^{\prime \prime }-x\left ( x+2\right ) y^{\prime }+\left ( x+2\right ) y & =0\\ x^{2}\left ( v^{\prime \prime }y_{1}+2v^{\prime }y_{1}^{\prime }+vy_{1}^{\prime \prime }\right ) -x\left ( x+2\right ) \left ( v^{\prime }y_{1}+vy_{1}^{\prime }\right ) +\left ( x+2\right ) \left ( vy_{1}\right ) & =0\\ v^{\prime \prime }\left ( x^{2}y_{1}\right ) +v^{\prime }\left ( 2x^{2}y_{1}^{\prime }-x\left ( x+2\right ) y_{1}\right ) +v\overset{0}{\overbrace{\left ( x^{2}y_{1}^{\prime \prime }-x\left ( x+2\right ) y_{1}^{\prime }+\left ( x+2\right ) y_{1}\right ) }} & =0 \end{align*}
Hence\[ v^{\prime \prime }\left ( x^{2}y_{1}\right ) +v^{\prime }\left ( 2x^{2}y_{1}^{\prime }-x\left ( x+2\right ) y_{1}\right ) =0 \] But \(y_{1}=x\), hence the above becomes\begin{align*} x^{3}v^{\prime \prime }+v^{\prime }\left ( 2x^{2}-x\left ( x+2\right ) x\right ) & =0\\ x^{3}v^{\prime \prime }-x^{3}v^{\prime } & =0 \end{align*}
Since we are told \(x>0\) when we can divide by \(x^{3}\) and obtain\[ v^{\prime \prime }-v^{\prime }=0 \] To solve the above, let \[ z=v^{\prime }\] Therefore \(z^{\prime }-z=0\) or \(\frac{d}{dx}\left ( ze^{x}\right ) =0\) or \(ze^{x}=c_{1}\) or \(z=c_{1}e^{-x}\). Therefore the above becomes\[ v^{\prime }=c_{1}e^{-x}\] Integrating\[ v=c_{2}-c_{1}e^{-x}\] Since \(y=vy_{1}\) therefore\[ y=y_{1}\left ( c_{2}-c_{1}e^{-x}\right ) \] But \(y_{1}=x\), hence the complete solution is\[ y=c_{2}x-c_{1}xe^{-x}\] Therefore, we see now that the two basis solutions are \begin{align*} y_{1} & =x\\ y_{2} & =xe^{x} \end{align*}
These can be shown to be L.I. using the Wronskian as follows
\begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} \\ & =\begin{vmatrix} x & xe^{x}\\ 1 & e^{x}+xe^{x}\end{vmatrix} \\ & =xe^{x}+x^{2}e^{x}-xe^{x}\\ & =x^{2}e^{x} \end{align*}
Which is not zero since we are told \(x>0\). Hence indeed the second basis solution \(y_{2}\) found is L.I. to \(y_{1}\).
Problem Find the particular solution for \(y^{\prime \prime }+2y^{\prime }-3y=1+xe^{x}\)
solution First we find the homogenous solution. This will tell us if \(e^{x}\) is one of the basis solutions of not, so we know what to guess. The characteristic equation is\begin{align*} r^{2}+2r-3 & =0\\ \left ( r-1\right ) \left ( r+3\right ) & =0 \end{align*}
Hence \(y_{1}=e^{x},y_{2}=e^{-3x}\). \(e^{x}\) is a solution to the homogeneous ODE. The guess is therefore\begin{align} y_{p} & =A+\left ( B+Cx\right ) xe^{x}\nonumber \\ & =A+\left ( Bx+Cx^{2}\right ) e^{x} \tag{1} \end{align}
Hence\begin{align*} y_{p}^{\prime } & =\left ( B+2Cx\right ) e^{x}+\left ( Bx+Cx^{2}\right ) e^{x}\\ & =e^{x}\left ( B+2Cx+Bx+Cx^{2}\right ) \\ y_{p}^{\prime \prime } & =\left ( 2C+B+2Cx\right ) e^{x}+e^{x}\left ( B+2Cx+Bx+Cx^{2}\right ) \\ & =e^{x}\left ( 2C+B+2Cx+B+2Cx+Bx+Cx^{2}\right ) \\ & =e^{x}\left ( 2C+2B+4Cx+Bx+Cx^{2}\right ) \end{align*}
Plugging into the ODE\begin{align*} e^{x}\left ( 2C+2B+4Cx+Bx+Cx^{2}\right ) +2e^{x}\left ( B+2Cx+Bx+Cx^{2}\right ) -3\left ( A+\left ( Bx+Cx^{2}\right ) e^{x}\right ) & =1+xe^{x}\\ e^{x}\left ( 2C+2B+2B\right ) +xe^{x}\left ( 4C+B+4C+2B-3B\right ) +x^{2}e^{x}\left ( C+2C-3C\right ) -3A & =1+xe^{x}\\ e^{x}\left ( 2C+4B\right ) +xe^{x}\left ( 8C\right ) -3A & =1+xe^{x} \end{align*}
Hence \(-3A=1\) or \(A=-\frac{1}{3}\) and \[ 8C=1 \] Or \[ C=\frac{1}{8}\] And \[ 2C+4B=0 \] Or \[ B=-\frac{1}{16}\] Hence particular solution becomes, from (1)\begin{align*} y_{p} & =-\frac{1}{3}+\left ( -\frac{1}{16}x+\frac{1}{8}x^{2}\right ) e^{x}\\ & =-\frac{1}{3}+\frac{1}{16}\left ( 2x^{2}-x\right ) e^{x} \end{align*}
Problem Find the particular solution for \(y^{\prime \prime }+9y=2\cos 3x+3\sin 3x\)
solution First we find the homogenous solution. The characteristic equation is\begin{align*} r^{2}+9 & =0\\ r^{2} & =-9\\ r & =\pm 3i \end{align*}
Hence \(y_{1}=e^{3ix},y_{2}=e^{-3ix}\) or \(y_{h}=c_{1}\cos 3x+c_{2}\sin 3x\). We see that \(\cos 3x\) and \(\sin 3x\) are already in the homogeneous solution. Therefore the guess is\[ y_{p}=Ax\cos 3x+Bx\sin 3x \] Hence\begin{align*} y_{p}^{\prime } & =A\cos 3x-3Ax\sin 3x+B\sin 3x+3Bx\cos 3x\\ y_{p}^{\prime \prime } & =-3A\sin 3x-3A\sin 3x-9Ax\cos 3x+3B\cos 3x+3B\cos 3x-9Bx\sin 3x \end{align*}
Substitution into the ODE gives\begin{multline*} \left ( -3A\sin 3x-3A\sin 3x-9Ax\cos 3x+3B\cos 3x+3B\cos 3x-9Bx\sin 3x\right ) \\ +9\left ( Ax\cos 3x+Bx\sin 3x\right ) =2\cos 3x+3\sin 3x \end{multline*} Or\begin{align*} -6A\sin 3x-9Ax\cos 3x+6B\cos 3x-9Bx\sin 3x+9Ax\cos 3x+9Bx\sin 3x & =2\cos 3x+3\sin 3x\\ \sin 3x\left ( -6A\right ) +\cos 3x\left ( 6B\right ) +x\sin 3x\left ( -9B+9B\right ) +x\cos 3x\left ( -9A+9A\right ) & =2\cos 3x+3\sin 3x\\ -6A\sin 3x+6B\cos 3x & =2\cos 3x+3\sin 3x \end{align*}
Hence \(-6A=3\) or \(A=\frac{-1}{2}\) and \(6B=2\) or \(B=\frac{1}{3}\), therefore the particular solution is\begin{align*} y_{p} & =\frac{-1}{2}x\cos 3x+\frac{1}{3}x\sin 3x\\ & =\frac{1}{6}\left ( 2x\sin 3x-3x\cos 3x\right ) \end{align*}
Problem Find the particular solution for \(y^{\prime \prime }+9y=2x^{2}e^{3x}+5\)
solution From the above problem, we found \(y_{h}=c_{1}\cos 3x+c_{2}\sin 3x\). Therefore there are no basis solutions in the RHS which are in the homogenous solution. The guess for the constant term is \(A\). The guess for \(2x^{2}e^{3x}\) is \(\left ( B_{0}+B_{1}x+B_{2}x^{2}\right ) e^{3x}\), hence\begin{align*} y_{p} & =A+\left ( B_{0}+B_{1}x+B_{2}x^{2}\right ) e^{3x}\\ y_{p}^{\prime } & =\left ( B_{1}+2B_{2}x\right ) e^{3x}+3\left ( B_{0}+B_{1}x+B_{2}x^{2}\right ) e^{3x}\\ y_{p}^{\prime \prime } & =2B_{2}e^{3x}+3\left ( B_{1}+2B_{2}x\right ) e^{3x}+3\left ( B_{1}+2B_{2}x\right ) e^{3x}+9\left ( B_{0}+B_{1}x+B_{2}x^{2}\right ) e^{3x} \end{align*}
Simplifying\begin{align*} y_{p}^{\prime \prime } & =e^{3x}\left ( 2B_{2}+3B_{1}+3B_{1}+9B_{0}\right ) +xe^{x}\left ( 6B_{2}+6B_{2}+9B_{1}\right ) +x^{2}e^{3x}\left ( 9B_{2}\right ) \\ & =e^{3x}\left ( 2B_{2}+6B_{1}+9B_{0}\right ) +xe^{x}\left ( 12B_{2}+9B_{1}\right ) +x^{2}e^{3x}\left ( 9B_{2}\right ) \end{align*}
Substitution into the ODE gives\begin{align*} e^{3x}\left ( 2B_{2}+6B_{1}+9B_{0}\right ) +xe^{x}\left ( 12B_{2}+9B_{1}\right ) +x^{2}e^{3x}\left ( 9B_{2}\right ) +9\left ( A+\left ( B_{0}+B_{1}x+B_{2}x^{2}\right ) e^{3x}\right ) & =2x^{2}e^{3x}+5\\ e^{3x}\left ( 2B_{2}+6B_{1}+9B_{0}\right ) +xe^{x}\left ( 12B_{2}+9B_{1}\right ) +x^{2}e^{3x}\left ( 9B_{2}\right ) +9A+\left ( 9B_{0}+9B_{1}x+9B_{2}x^{2}\right ) e^{3x} & =2x^{2}e^{3x}+5\\ e^{3x}\left ( 2B_{2}+6B_{1}+18B_{0}\right ) +xe^{x}\left ( 12B_{2}+18B_{1}\right ) +x^{2}e^{3x}\left ( 18B_{2}\right ) +9A & =2x^{2}e^{3x}+5 \end{align*}
Comparing coefficients gives\begin{align*} 9A & =5\\ 2B_{2}+6B_{1}+18B_{0} & =0\\ 12B_{2}+18B_{1} & =0\\ 19B_{2} & =2 \end{align*}
From last equation \(B_{2}=\frac{1}{9}\). Hence from third equation \(18B_{1}=-\frac{12}{9}\), or \(B_{1}=-\frac{2}{27}\). And from second equation \begin{align*} 2B_{2}+6B_{1}+18B_{0} & =0\\ 2\left ( \frac{1}{9}\right ) +6\left ( -\frac{2}{27}\right ) +18B_{0} & =0\\ B_{0} & =\frac{1}{81} \end{align*}
And \(A=\frac{5}{9}\). Therefore \begin{align*} y_{p} & =A+\left ( B_{0}+B_{1}x+B_{2}x^{2}\right ) e^{3x}\\ & =\frac{5}{9}+\left ( \frac{1}{81}-\frac{2}{27}x+\frac{1}{9}x^{2}\right ) e^{3x}\\ & =\frac{5}{9}+\left ( \frac{1}{81}-\frac{6}{81}x+\frac{9}{81}x^{2}\right ) e^{3x}\\ & =\frac{45}{81}+\left ( \frac{1}{81}-\frac{6}{81}x+\frac{9}{81}x^{2}\right ) e^{3x}\\ & =\frac{1}{81}\left ( 45+e^{3x}-6xe^{x}+9x^{2}e^{3x}\right ) \end{align*}
Problem Setup the form for the particular solution but do not determine the values of the coefficients. \(y^{\prime \prime }+3y^{\prime }+2y=xe^{-x}-xe^{-2x}\)
solution First we find the homogenous solution. The characteristic equation is\begin{align*} r^{2}+3r+2 & =0\\ \left ( r+1\right ) \left ( r+2\right ) & =0 \end{align*}
Hence \(y_{1}=e^{-x},y_{2}=e^{-2x}\). We see that the basis solutions are part of the RHS. Therefore the guess solution is\[ y_{p}=x\left ( A_{1}+A_{2}x\right ) e^{-x}+x\left ( A_{3}+A_{4}x\right ) e^{-2x}\]
Problem Setup the form for the particular solution but do not determine the values of the coefficients. \(y^{\prime \prime }-6y^{\prime }+13y=xe^{3x}\sin 2x\)
solution First we find the homogenous solution. The characteristic equation is\[ r^{2}-6r+13=0 \] The roots are \(3\pm 2i\). Hence the homogenous solution is \(y_{h}=c_{1}e^{3x}\cos 2x+c_{2}e^{3x}\sin 2x\). We see that \(e^{3x}\sin 2x\) is already in the homogenous solution. Hence the guess is \begin{align*} y_{p} & =\overset{x\text{ guess}}{\overbrace{\left ( A_{1}+A_{2}x\right ) }}x\overset{\sin 2xe^{3x}\text{ guess}}{\overbrace{\left ( A_{3}\sin 2x+A_{4}\cos 2x\right ) e^{3x}}}\\ & =\left ( A_{1}x+A_{2}x^{2}\right ) e^{3x}\cos 2x+\left ( A_{3}x+A_{4}x^{2}\right ) e^{3x}\sin 2x \end{align*}
Problem Solve the initial value problem \(y^{\prime \prime \prime }-2y^{\prime \prime }+y^{\prime }=1+xe^{x}\) with \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0,y^{\prime \prime }\left ( 0\right ) =1\)
solution First we find the homogenous solution. The characteristic equation is\begin{align*} r^{3}-2r^{2}+r & =0\\ r\left ( r^{2}-2r+1\right ) & =0 \end{align*}
For \(r^{2}-2r+1=0\), it factors into \(\left ( r-1\right ) \left ( r-1\right ) \), hence roots are \(r_{1}=0,r_{2}=1,r_{3}=1\). Since double roots, the homogenous solution is\[ y_{h}=c_{1}+c_{2}e^{x}+c_{3}xe^{x}\] We notice that both \(e^{x}\) and \(xe^{x}\) is in the RHS. Therefore we need to multiply by \(x^{2}\). The guess is therefore\begin{align*} y_{p} & =Ax+x^{2}\left ( B+Cx\right ) e^{x}\\ & =Ax+\left ( Bx^{2}+Cx^{3}\right ) e^{x} \end{align*}
Therefore\begin{align*} y_{p}^{\prime } & =A+\left ( 2Bx+3Cx^{2}\right ) e^{x}+\left ( Bx^{2}+Cx^{3}\right ) e^{x}\\ y_{p}^{\prime \prime } & =\left ( 2B+6Cx\right ) e^{x}+\left ( 2Bx+3Cx^{2}\right ) e^{x}+\left ( 2Bx+3Cx^{2}\right ) e^{x}+\left ( Bx^{2}+Cx^{3}\right ) e^{x} \end{align*}
Simplifying gives\begin{align*} y_{p}^{\prime } & =A+xe^{x}\left ( 2B\right ) +x^{2}e^{x}\left ( 3C+B\right ) +x^{3}e^{x}\left ( C\right ) \\ y_{p}^{\prime \prime } & =e^{x}\left ( 2B\right ) +xe^{x}\left ( 6C+4B\right ) +x^{2}e^{x}\left ( 6C+B\right ) +x^{3}e^{x}\left ( C\right ) \\ y_{p}^{\prime \prime \prime } & =e^{x}\left ( 2B\right ) +e^{x}\left ( 6C+4B\right ) +xe^{x}\left ( 6C+4B\right ) +2xe^{x}\left ( 6C+B\right ) +x^{2}e^{x}\left ( 6C+B\right ) +3x^{2}e^{x}\left ( C\right ) +x^{3}e^{x}\left ( C\right ) \\ & =e^{x}\left ( 6B+6C\right ) +xe^{x}\left ( 6C+4B+12C+2B\right ) +x^{2}e^{x}\left ( 6C+B+3C\right ) +Cx^{3}e^{x}\\ & =e^{x}\left ( 6B+6C\right ) +xe^{x}\left ( 18C+6B\right ) +x^{2}e^{x}\left ( 9C+B\right ) +Cx^{3}e^{x} \end{align*}
Substitution into the ODE gives\[ y_{p}^{\prime \prime \prime }-2y_{p}^{\prime \prime }+y_{p}^{\prime }=1+xe^{x}\] Hence\begin{multline*} e^{x}\left ( 6B+6C\right ) +xe^{x}\left ( 18C+6B\right ) +x^{2}e^{x}\left ( 9C+B\right ) +Cx^{3}e^{x}\\ -2\left ( e^{x}\left ( 2B\right ) +xe^{x}\left ( 6C+4B\right ) +x^{2}e^{x}\left ( 6C+B\right ) +x^{3}e^{x}\left ( C\right ) \right ) +\\ A+xe^{x}\left ( 2B\right ) +x^{2}e^{x}\left ( 3C+B\right ) +x^{3}e^{x}\left ( C\right ) =1+xe^{x} \end{multline*} Or\begin{multline*} e^{x}\left ( 6B+6C\right ) +xe^{x}\left ( 18C+6B\right ) +x^{2}e^{x}\left ( 9C+B\right ) +Cx^{3}e^{x}\\ -e^{x}\left ( 4B\right ) -xe^{x}\left ( 12C+8B\right ) -x^{2}e^{x}\left ( 12C+2B\right ) -x^{3}e^{x}\left ( 2C\right ) +\\ A+xe^{x}\left ( 2B\right ) +x^{2}e^{x}\left ( 3C+B\right ) +x^{3}e^{x}\left ( C\right ) =1+xe^{x} \end{multline*} Or\begin{multline*} e^{x}\left ( 6B+6C-4B\right ) +xe^{x}\left ( 18C+6B-12C-8B+2B\right ) +\\ x^{2}e^{x}\left ( 9C+B-12C-2B+3C+B\right ) +x^{3}e^{x}\left ( C-2C+C\right ) +A=1+xe^{x} \end{multline*} Or\[ e^{x}\left ( 2B+6C\right ) +xe^{x}\left ( 6C\right ) +A=1+xe^{x}\] Hence\begin{align*} 6C & =1\\ 2B+6C & =0\\ A & =1 \end{align*}
Therefore, \(C=\frac{1}{6},B=-\frac{1}{2}\), and the particular solution is\begin{align*} y_{p} & =Ax+x^{2}\left ( B+Cx\right ) e^{x}\\ & =x+\left ( -\frac{1}{2}x^{2}+\frac{1}{6}x^{3}\right ) e^{x} \end{align*}
Hence the complete solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}+c_{2}e^{x}+c_{3}xe^{x}+x+\left ( -\frac{1}{2}x^{2}+\frac{1}{6}x^{3}\right ) e^{x} \end{align*}
Applying initial conditions. \(y\left ( 0\right ) =0\) gives\begin{equation} 0=c_{1}+c_{2} \tag{1} \end{equation} And \[ y^{\prime }=c_{2}e^{x}+c_{3}e^{x}+c_{3}xe^{x}+1+\left ( -x+\frac{1}{2}x^{2}\right ) e^{x}+\left ( -\frac{1}{2}x^{2}+\frac{1}{6}x^{3}\right ) e^{x}\] Applying second initial conditions \(y^{\prime }\left ( 0\right ) =0\) gives\begin{equation} 0=c_{2}+c_{3}+1 \tag{2} \end{equation} And \[ y^{\prime \prime }=c_{2}e^{x}+c_{3}e^{x}+c_{3}e^{x}+c_{3}xe^{x}+\left ( -1+x\right ) e^{x}+\left ( -x+\frac{1}{2}x^{2}\right ) e^{x}+\left ( -x+\frac{1}{2}x^{2}\right ) e^{x}+\left ( -\frac{1}{2}x^{2}+\frac{1}{6}x^{3}\right ) e^{x}\] Applying initial conditions \(y^{\prime \prime }\left ( 0\right ) =1\) gives\begin{align*} 1 & =c_{2}+2c_{3}-1\\ 2 & =c_{2}+2c_{3} \end{align*}
The solution is \(c_{1}=4,c_{2}=-4,c_{3}=3\), hence the general solution is\begin{align*} y & =c_{1}+c_{2}e^{x}+c_{3}xe^{x}+x+\left ( -\frac{1}{2}x^{2}+\frac{1}{6}x^{3}\right ) e^{x}\\ & =4-4e^{x}+3xe^{x}+x-\frac{1}{2}x^{2}e^{x}+\frac{1}{6}x^{3}e^{x} \end{align*}
Problem Use method of variation of parameters to find particular solution \(y^{\prime \prime }-4y^{\prime }+4y=2e^{2x}\)
solution We need to first find the homogenous solution. The characteristic equation is\begin{align*} r^{2}-4r+4 & =0\\ \left ( r-2\right ) \left ( r-2\right ) & =0 \end{align*}
Hence \(r_{1}=2\), double root. Therefore \begin{align*} y_{1}\left ( x\right ) & =e^{2x}\\ y_{2}\left ( x\right ) & =xe^{2x} \end{align*}
Let \[ y_{p}=u_{1}y_{1}+u_{2}y_{2}\] Where\begin{align*} u_{1} & =-\int \frac{y_{2}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx\\ u_{2} & =\int \frac{y_{1}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx \end{align*}
Where \(f\left ( x\right ) =2e^{2x}\) and \begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{2x} & xe^{2x}\\ 2e^{2x} & e^{2x}+2xe^{2x}\end{vmatrix} \\ & =e^{2x}\left ( e^{2x}+2xe^{2x}\right ) -2xe^{4x}\\ & =e^{4x}+2xe^{4x}-2xe^{4x}\\ & =e^{4x} \end{align*}
Hence\[ u_{1}=-\int \frac{xe^{2x}\left ( 2e^{2x}\right ) }{e^{4x}}dx=-\int 2xdx=-x^{2}\] And\[ u_{2}=\int \frac{e^{2x}\left ( 2e^{2x}\right ) }{e^{4x}}dx=2x \] Therefore\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =-x^{2}e^{2x}+2x^{2}e^{2x}\\ & =x^{2}e^{2x} \end{align*}
Problem Use method of variation of parameters to find particular solution \(y^{\prime \prime }-4y=\sinh 2x\)
solution We need to first find the homogenous solution. The characteristic equation is\begin{align*} r^{2}-4 & =0\\ r & =\pm 2 \end{align*}
Therefore \begin{align*} y_{1}\left ( x\right ) & =e^{2x}\\ y_{2}\left ( x\right ) & =e^{-2x} \end{align*}
Let \[ y_{p}=u_{1}y_{1}+u_{2}y_{2}\] Where\begin{align*} u_{1} & =-\int \frac{y_{2}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx\\ u_{2} & =\int \frac{y_{1}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx \end{align*}
Where \(f\left ( x\right ) =\sinh 2x=\frac{e^{2x}-e^{-2x}}{2}\) and \begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} e^{2x} & e^{-2x}\\ 2e^{2x} & -2e^{-2x}\end{vmatrix} \\ & =-2-2=-4 \end{align*}
Hence\begin{align*} u_{1} & =-\int \frac{e^{-2x}\left ( \frac{e^{2x}-e^{-2x}}{2}\right ) }{-4}dx\\ & =\frac{1}{4}\int e^{-2x}\left ( \frac{e^{2x}-e^{-2x}}{2}\right ) dx\\ & =\frac{1}{8}\int \left ( 1-e^{-4x}\right ) dx\\ & =\frac{1}{8}\left ( x+\frac{e^{-4x}}{4}\right ) \end{align*}
And\begin{align*} u_{2} & =\int \frac{e^{2x}\left ( \frac{e^{2x}-e^{-2x}}{2}\right ) }{-4}dx\\ & =-\frac{1}{8}\int e^{2x}\left ( e^{2x}-e^{-2x}\right ) dx\\ & =-\frac{1}{8}\int \left ( e^{4x}-1\right ) dx\\ & =-\frac{1}{8}\left ( \frac{e^{4x}}{4}-x\right ) \end{align*}
Therefore\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\frac{1}{8}\left ( x+\frac{e^{-4x}}{4}\right ) e^{2x}-\frac{1}{8}\left ( \frac{e^{4x}}{4}-x\right ) e^{-2x}\\ & =\left ( \frac{1}{8}xe^{2x}+\frac{e^{-2x}}{32}\right ) -\frac{1}{8}\left ( \frac{e^{2x}}{32}-\frac{xe^{-2x}}{8}\right ) \\ & =\frac{1}{8}xe^{2x}+\frac{e^{-2x}}{32}-\frac{e^{2x}}{32}+\frac{xe^{-2x}}{8}\\ & =\frac{1}{4}x\left ( \frac{e^{2x}+e^{-2x}}{2}\right ) +\frac{1}{16}\left ( \frac{e^{-2x}-e^{2x}}{2}\right ) \\ & =\frac{1}{4}x\left ( \frac{e^{2x}+e^{-2x}}{2}\right ) -\frac{1}{16}\left ( \frac{e^{2x}-e^{-2x}}{2}\right ) \\ & =\frac{1}{4}x\cosh 2x-\frac{1}{16}\sinh 2x\\ & =\frac{1}{16}\left ( 4x\cosh 2x-\sinh 2x\right ) \end{align*}
Problem Use method of variation of parameters to find particular solution \(y^{\prime \prime }+9y=2\sec 3x\)
solution We need to first find the homogenous solution. The characteristic equation is\begin{align*} r^{2}+9 & =0\\ r & =\pm 3i \end{align*}
Therefore \begin{align*} y_{1}\left ( x\right ) & =\sin 3x\\ y_{2}\left ( x\right ) & =\cos 3x \end{align*}
Let \[ y_{p}=u_{1}y_{1}+u_{2}y_{2}\] Where\begin{align*} u_{1} & =-\int \frac{y_{2}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx\\ u_{2} & =\int \frac{y_{1}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx \end{align*}
Where \(f\left ( x\right ) =2\sec 3x=\frac{2}{\cos 3x}\) and \begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} \sin 3x & \cos 3x\\ 3\cos 3x & -3\sin 3x \end{vmatrix} \\ & =-3\sin ^{2}3x-3\cos ^{2}x\\ & =-3 \end{align*}
Hence\begin{align*} u_{1} & =-\int \frac{\cos 3x\left ( \frac{2}{\cos 3x}\right ) }{-3}dx\\ & =\frac{1}{3}\int 2dx\\ & =\frac{2}{3}x \end{align*}
And\begin{align*} u_{2} & =\int \frac{\sin 3x\left ( \frac{2}{\cos 3x}\right ) }{-3}dx\\ & =\frac{-2}{3}\int \tan 3xdx\\ & =\frac{-2}{3}\left ( \frac{1}{6}\ln \frac{1}{\cos ^{2}\left ( 3x\right ) }\right ) \end{align*}
Therefore\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\frac{2}{3}x\left ( \sin 3x\right ) +\frac{-2}{3}\left ( \frac{1}{6}\ln \frac{1}{\cos ^{2}\left ( 3x\right ) }\right ) \cos 3x\\ & =\frac{2}{3}x\left ( \sin 3x\right ) -\frac{1}{9}\cos \left ( 3x\right ) \ln \left ( \frac{1}{\cos ^{2}\left ( 3x\right ) }\right ) \\ & =\frac{2}{3}x\left ( \sin 3x\right ) +\frac{1}{9}\cos \left ( 3x\right ) \ln \left ( \cos ^{2}\left ( 3x\right ) \right ) \\ & =\frac{2}{3}x\left ( \sin 3x\right ) +\frac{2}{9}\cos \left ( 3x\right ) \ln \left \vert \cos \left ( 3x\right ) \right \vert \end{align*}
Problem Find a particular solution to the Euler ODE \(x^{2}y^{\prime \prime }+xy^{\prime }+y=\ln x\) with homogenous solution \(y_{h}=c_{1}\cos \left ( \ln x\right ) +c_{2}\sin \left ( \ln x\right ) \)
solution We see that \begin{align*} y_{1} & =\cos \left ( \ln x\right ) \\ y_{2} & =\sin \left ( \ln x\right ) \end{align*}
Using variation of parameters on the ODE\[ y^{\prime \prime }+\frac{1}{x}y^{\prime }+\frac{1}{x^{2}}y=\frac{\ln x}{x^{2}}\] Where now we use \(f\left ( x\right ) =\frac{\ln x}{x^{2}}\). Let \[ y_{p}=u_{1}y_{1}+u_{2}y_{2}\] Where\begin{align*} u_{1} & =-\int \frac{y_{2}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx\\ u_{2} & =\int \frac{y_{1}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx \end{align*}
And \begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} \cos \left ( \ln x\right ) & \sin \left ( \ln x\right ) \\ -\frac{1}{x}\sin \left ( \ln x\right ) & \frac{1}{x}\cos \left ( \ln x\right ) \end{vmatrix} \\ & =\frac{1}{x}\cos ^{2}\left ( \ln x\right ) +\frac{1}{x}\sin ^{2}\left ( \ln x\right ) \\ & =\frac{1}{x} \end{align*}
Hence\begin{align*} u_{1} & =-\int \frac{\sin \left ( \ln x\right ) \left ( \frac{\ln x}{x^{2}}\right ) }{\frac{1}{x}}dx\\ & =-\int \frac{\ln x\sin \left ( \ln x\right ) }{x}dx\\ & =\ln \left ( x\right ) \cos \left ( \ln x\right ) -\sin \left ( \ln x\right ) \end{align*}
And\begin{align*} u_{2} & =\int \frac{\cos \left ( \ln x\right ) \left ( \frac{\ln x}{x^{2}}\right ) }{\frac{1}{x}}dx\\ & =\int \frac{\cos \left ( \ln x\right ) \left ( \ln x\right ) }{x}dx\\ & =\ln \left ( x\right ) \sin \left ( \ln x\right ) +\cos \left ( \ln x\right ) \end{align*}
Therefore\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( \ln \left ( x\right ) \cos \left ( \ln x\right ) -\sin \left ( \ln x\right ) \right ) \cos \left ( \ln x\right ) +\left ( \ln \left ( x\right ) \sin \left ( \ln x\right ) +\cos \left ( \ln x\right ) \right ) \sin \left ( \ln x\right ) \\ & =\ln \left ( x\right ) \cos ^{2}\left ( \ln x\right ) -\sin \left ( \ln x\right ) \cos \left ( \ln x\right ) +\ln \left ( x\right ) \sin ^{2}\left ( \ln x\right ) +\sin \left ( \ln \right ) \cos \left ( \ln x\right ) \\ & =\ln \left ( x\right ) \cos ^{2}\left ( \ln x\right ) +\ln \left ( x\right ) \sin ^{2}\left ( \ln x\right ) \\ & =\ln x \end{align*}
Problem Find a particular solution to the Euler ODE \(\left ( x^{2}-1\right ) y^{\prime \prime }-2xy^{\prime }+2y=x^{2}-1\) with homogenous solution \(y_{h}=c_{1}x+c_{2}\left ( 1+x^{2}\right ) \)
solution We see that \begin{align*} y_{1} & =x\\ y_{2} & =1+x^{2} \end{align*}
Using variation of parameters on the ODE\[ y^{\prime \prime }-2\frac{x}{\left ( x^{2}-1\right ) }y^{\prime }+\frac{2}{\left ( x^{2}-1\right ) }y=1 \] Where now we use \(f\left ( x\right ) =1\). Let \[ y_{p}=u_{1}y_{1}+u_{2}y_{2}\] Where\begin{align*} u_{1} & =-\int \frac{y_{2}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx\\ u_{2} & =\int \frac{y_{1}\left ( x\right ) f\left ( x\right ) }{W\left ( x\right ) }dx \end{align*}
And \begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} x & 1+x^{2}\\ 1 & 2x \end{vmatrix} \\ & =2x^{2}-\left ( 1+x^{2}\right ) \\ & =x^{2}-1 \end{align*}
Hence\begin{align*} u_{1} & =-\int \frac{\left ( 1+x^{2}\right ) \left ( 1\right ) }{x^{2}-1}dx\\ & =-x-\ln \left ( x-1\right ) +\ln \left ( x+1\right ) \end{align*}
And\begin{align*} u_{2} & =\int \frac{x}{x^{2}-1}dx\\ & =\frac{1}{2}\ln \left ( x-1\right ) +\frac{1}{2}\ln \left ( x+1\right ) \end{align*}
Therefore\begin{align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\left ( -x-\ln \left ( x-1\right ) +\ln \left ( x+1\right ) \right ) x+\left ( \frac{1}{2}\ln \left ( x-1\right ) +\frac{1}{2}\ln \left ( x+1\right ) \right ) \left ( 1+x^{2}\right ) \\ & =-x^{2}+x\ln \left \vert \frac{x+1}{x-1}\right \vert +\frac{1}{2}\left ( 1+x^{2}\right ) \ln \left \vert \left ( x-1\right ) \left ( x+1\right ) \right \vert \\ & =-x^{2}+x\ln \left \vert \frac{x+1}{x-1}\right \vert +\frac{1}{2}\left ( 1+x^{2}\right ) \ln \left \vert x^{2}-1\right \vert \end{align*}