This help page gives a few examples of using the command ODESteps to solve second order initial value problems.

See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right)\:$

$\mathrm{ivp1}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-\mathrm{diff}\left(y\left(x\right)\,x\right)-x\exp \left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=0\right)=0\,y\left(0\right)=1\right\}$

$\mathrm{ivp1}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\,y\left(0\right)=1\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\,y\left(0\right)=1\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)+x{\ⅇ}^x\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)=x{\ⅇ}^x\\ \text{•}& & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2-r=0\\ \text{•}& & \text{Factor the characteristic polynomial}\\ & & r\left(r-1\right)=0\\ \text{•}& & \text{Roots of the characteristic polynomial}\\ & & r=\left(0\,1\right)\\ \text{•}& & \text{1st solution of the homogeneous ODE}\\ & & y_1\left(x\right)=1\\ \text{•}& & \text{2nd solution of the homogeneous ODE}\\ & & y_2\left(x\right)={\ⅇ}^x\\ \text{•}& & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathrm{\_C1}{y}_1\left(x\right)+\mathrm{\_C2}{y}_2\left(x\right)+y_p\left(x\right)\\ \text{•}& & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(x\right)=\mathrm{\_C1}+\mathrm{\_C2}{\ⅇ}^x+y_p\left(x\right)\\ \text{▫}& & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & \text{◦}& \text{Use variation of paramaters to find}{y}_p\text{here}f\left(x\right)\text{is the forcing function}\\ & & \left[y_p\left(x\right)=-y_1\left(x\right)\left(\int \frac{y_2\left(x\right)f\left(x\right)}{W\left(y_1\left(x\right)\,y_2\left(x\right)\right)}\ⅆx\right)+y_2\left(x\right)\left(\int \frac{y_1\left(x\right)f\left(x\right)}{W\left(y_1\left(x\right)\,y_2\left(x\right)\right)}\ⅆx\right)\,f\left(x\right)=x{\ⅇ}^x\right]\\ & \text{◦}& \text{Wronskian of solutions of the homogeneous equation}\\ & & W\left(y_1\left(x\right)\,y_2\left(x\right)\right)=\left[\begin{array}{cc}1& {\ⅇ}^x\\ 0& {\ⅇ}^x\end{array}\right]\\ & \text{◦}& \text{Compute Wronskian}\\ & & W\left(y_1\left(x\right)\,y_2\left(x\right)\right)={\ⅇ}^x\\ & \text{◦}& \text{Substitute functions into equation for}{y}_p\left(x\right)\\ & & y_p\left(x\right)=-\left(\int x{\ⅇ}^x\ⅆx\right)+{\ⅇ}^x\left(\int x\ⅆx\right)\\ & \text{◦}& \text{Compute integrals}\\ & & y_p\left(x\right)=\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\\ \text{•}& & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(x\right)=\mathrm{\_C1}+\mathrm{\_C2}{\ⅇ}^x+\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\\ \text{•}& & \text{Use initial condition}y\left(0\right)=1\\ & & 1=\mathrm{\_C1}+\mathrm{\_C2}+1\\ \text{•}& & \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\mathrm{\_C2}{\ⅇ}^x+\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}+\frac{{\ⅇ}^x\left(2x-2\right)}{2}\\ \text{•}& & \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\\ & & 0=\mathrm{\_C2}\\ \text{•}& & \text{Solve for}\mathrm{\_C1}\text{and}\mathrm{\_C2}\\ & & \left\{\mathrm{\_C1}=0\,\mathrm{\_C2}=0\right\}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\end{array}$

$\mathrm{ivp2}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+\frac{5{\mathrm{diff}\left(y\left(x\right)\,x\right)}^2}{y\left(x\right)}=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=1\right)=-3\,y\left(1\right)=1\right\}$

$\mathrm{ivp2}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\,y\left(1\right)=1\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{5{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2}{y\left(x\right)}=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\,y\left(1\right)=1\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Define new dependent variable}y\\ & & u\left(x\right)=\left[\right]\\ \text{•}& & \text{Compute}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & & \left[\right]=\left[\right]\\ \text{•}& & \text{Use chain rule on the lhs}\\ & & \left(\left[\right]\right)\left(\left[\right]\right)=\left[\right]\\ \text{•}& & \text{Substitute in the definition of}u\\ & & u\left(y\right)\left(\left[\right]\right)=\left[\right]\\ \text{•}& & \text{Make substitutions}\frac{\ⅆ}{\ⅆx}y\left(x\right)=u\left(y\right)\text{,}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)\text{to reduce order of ODE}\\ & & u\left(y\right)\left(\frac{\ⅆ}{\ⅆy}u\left(y\right)\right)+\frac{5{u\left(y\right)}^2}{y}=0\\ \text{•}& & \text{Separate variables}\\ & & \frac{\frac{\ⅆ}{\ⅆy}u\left(y\right)}{u\left(y\right)}=-\frac{5}{y}\\ \text{•}& & \text{Integrate both sides with respect to}y\\ & & \int \frac{\frac{\ⅆ}{\ⅆy}u\left(y\right)}{u\left(y\right)}\ⅆy=\int -\frac{5}{y}\ⅆy+\mathrm{\_C1}\\ \text{•}& & \text{Evaluate integral}\\ & & \ln \left(u\left(y\right)\right)=-5\ln \left(y\right)+\mathrm{\_C1}\\ \text{•}& & \text{Solve for}u\left(y\right)\\ & & u\left(y\right)=\frac{{\ⅇ}^{\mathrm{\_C1}}}{y^5}\\ \text{•}& & \text{Solve 1st ODE for}y\left(y\right)\\ & & u\left(y\right)=\frac{{\ⅇ}^{\mathrm{\_C1}}}{y^5}\\ \text{•}& & \text{Revert to original variables with substitution}u\left(y\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{,}y=y\left(x\right)\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{{\ⅇ}^{\mathrm{\_C1}}}{{y\left(x\right)}^5}\\ \text{•}& & \text{Separate variables}\\ & & \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right){y\left(x\right)}^5={\ⅇ}^{\mathrm{\_C1}}\\ \text{•}& & \text{Integrate both sides with respect to}x\\ & & \int \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right){y\left(x\right)}^5\ⅆx=\int {\ⅇ}^{\mathrm{\_C1}}\ⅆx+\mathrm{\_C2}\\ \text{•}& & \text{Evaluate integral}\\ & & \frac{{y\left(x\right)}^6}{6}={\ⅇ}^{\mathrm{\_C1}}x+\mathrm{\_C2}\\ \text{•}& & \text{Solve for}y\left(x\right)\\ & & \left\{y\left(x\right)={\left(6{\ⅇ}^{\mathrm{\_C1}}x+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\,y\left(x\right)=-{\left(6{\ⅇ}^{\mathrm{\_C1}}x+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right\}\\ \text{•}& & \text{Use initial condition}y\left(1\right)=1\\ & & 1={\left(6{\ⅇ}^{\mathrm{\_C1}}+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ \text{•}& & \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{{\ⅇ}^{\mathrm{\_C1}}}{{\left(6{\ⅇ}^{\mathrm{\_C1}}x+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ \text{•}& & \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\\ & & \mathrm{-3}=\frac{{\ⅇ}^{\mathrm{\_C1}}}{{\left(6{\ⅇ}^{\mathrm{\_C1}}+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ \text{•}& & \text{Solution is complex}\\ \text{•}& & \text{Use initial condition}y\left(1\right)=1\\ & & 1=-{\left(6{\ⅇ}^{\mathrm{\_C1}}+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ \text{•}& & \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{{\ⅇ}^{\mathrm{\_C1}}}{{\left(6{\ⅇ}^{\mathrm{\_C1}}x+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ \text{•}& & \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=\mathrm{-3}\\ & & \mathrm{-3}=-\frac{{\ⅇ}^{\mathrm{\_C1}}}{{\left(6{\ⅇ}^{\mathrm{\_C1}}+6\mathrm{\_C2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ \text{•}& & \text{Solution is complex}\end{array}$

$\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-\mathrm{diff}\left(y\left(x\right)\,x\right)-6y\left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=1\right)=a\,y\left(1\right)=0\right\}$

$\mathrm{ivp3}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-6y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=a\,y\left(1\right)=0\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-6y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=a\,y\left(1\right)=0\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Characteristic polynomial of ODE}\\ & & r^2-r-6=0\\ \text{•}& & \text{Factor the characteristic polynomial}\\ & & \left(r+2\right)\left(r-3\right)=0\\ \text{•}& & \text{Roots of the characteristic polynomial}\\ & & r=\left(\mathrm{-2}\,3\right)\\ \text{•}& & \text{1st solution of the ODE}\\ & & y_1\left(x\right)={\ⅇ}^{-2x}\\ \text{•}& & \text{2nd solution of the ODE}\\ & & y_2\left(x\right)={\ⅇ}^{3x}\\ \text{•}& & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathrm{\_C1}{y}_1\left(x\right)+\mathrm{\_C2}{y}_2\left(x\right)\\ \text{•}& & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathrm{\_C1}{\ⅇ}^{-2x}+\mathrm{\_C2}{\ⅇ}^{3x}\\ \text{•}& & \text{Use initial condition}y\left(1\right)=0\\ & & 0=\mathrm{\_C1}{\ⅇ}^{\mathrm{-2}}+\mathrm{\_C2}{\ⅇ}^3\\ \text{•}& & \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-2\mathrm{\_C1}{\ⅇ}^{-2x}+3\mathrm{\_C2}{\ⅇ}^{3x}\\ \text{•}& & \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=a\\ & & a=-2\mathrm{\_C1}{\ⅇ}^{\mathrm{-2}}+3\mathrm{\_C2}{\ⅇ}^3\\ \text{•}& & \text{Solve for}\mathrm{\_C1}\text{and}\mathrm{\_C2}\\ & & \left\{\mathrm{\_C1}=-\frac{a}{5{\ⅇ}^{\mathrm{-2}}}\,\mathrm{\_C2}=\frac{a}{5{\ⅇ}^3}\right\}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=-\frac{a{\ⅇ}^{-2x}}{5{\ⅇ}^{\mathrm{-2}}}+\frac{a{\ⅇ}^{3x}}{5{\ⅇ}^3}\end{array}$

$\mathrm{ivp4}≔\left\{x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)-4x\mathrm{diff}\left(y\left(x\right)\,x\right)+2y\left(x\right)=0\,\mathrm{eval}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x=1\right)=10\,y\left(1\right)=-1\right\}$

$\mathrm{ivp4}≔\left\{x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=10\,y\left(1\right)=\mathrm{-1}\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp4}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=10\,y\left(1\right)=\mathrm{-1}\right\}\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=-\frac{2y\left(x\right)}{x^2}+\frac{4\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x}\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{4\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x}+\frac{2y\left(x\right)}{x^2}=0\\ \text{•}& & \text{Multiply by denominators of the ODE}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-4x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2y\left(x\right)=0\\ \text{•}& & \text{Make a change of variables}\\ & & t=\ln \left(x\right)\\ \text{▫}& & \text{Substitute the change of variables back into the ODE}\\ & \text{◦}& \text{Calculate the 1st derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{\ⅆ}{\ⅆt}y\left(t\right)=\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x}\\ & \text{◦}& \text{Calculate the 2nd derivative of}y\text{with respect to}x\text{, using the chain rule}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ & \text{◦}& \text{Compute derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & x^2\left(-\frac{\frac{\ⅆ}{\ⅆt}y\left(t\right)}{x^2}+\frac{\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)}{x^2}\right)-4\frac{\ⅆ}{\ⅆt}y\left(t\right)+2y\left(t\right)=0\\ \text{•}& & \text{Simplify}\\ & & -5\frac{\ⅆ}{\ⅆt}y\left(t\right)+\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)+2y\left(t\right)=0\\ \text{•}& & \text{Characteristic polynomial of ODE}\\ & & r^2-5r+2=0\\ \text{•}& & \text{Use quadratic formula to solve for}r\\ & & r=\frac{5\pm \left(\left[\right]\right)}{2}\\ \text{•}& & \text{Roots of the characteristic polynomial}\\ & & r=\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\,\frac{5}{2}+\frac{\sqrt{17}}{2}\right)\\ \text{•}& & \text{1st solution of the ODE}\\ & & y_1\left(t\right)={\ⅇ}^{\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)t}\\ \text{•}& & \text{2nd solution of the ODE}\\ & & y_2\left(t\right)={\ⅇ}^{\left(\frac{5}{2}+\frac{\sqrt{17}}{2}\right)t}\\ \text{•}& & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathrm{\_C1}{y}_1\left(t\right)+\mathrm{\_C2}{y}_2\left(t\right)\\ \text{•}& & \text{Substitute in solutions}\\ & & y\left(t\right)=\mathrm{\_C1}{\ⅇ}^{\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)t}+\mathrm{\_C2}{\ⅇ}^{\left(\frac{5}{2}+\frac{\sqrt{17}}{2}\right)t}\\ \text{•}& & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\mathrm{\_C1}{\ⅇ}^{\left(\frac{5}{2}-\frac{\sqrt{17}}{2}\right)\left[\right]}+\mathrm{\_C2}{\ⅇ}^{\left(\frac{5}{2}+\frac{\sqrt{17}}{2}\right)\left[\right]}\\ \text{•}& & \text{Simplify}\\ & & y\left(x\right)=\mathrm{\_C1}{\ⅇ}^{-\frac{\left(-5+\sqrt{17}\right)\left[\right]}{2}}+\mathrm{\_C2}{\ⅇ}^{\frac{\left(5+\sqrt{17}\right)\left[\right]}{2}}\\ \text{•}& & \text{Use initial condition}y\left(1\right)=\mathrm{-1}\\ & & \mathrm{-1}=\mathrm{\_C1}+\mathrm{\_C2}\\ \text{•}& & \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{\mathrm{\_C1}\left(-5+\sqrt{17}\right){\ⅇ}^{-\frac{\left(-5+\sqrt{17}\right)\ln \left(x\right)}{2}}}{2x}+\frac{\mathrm{\_C2}\left(5+\sqrt{17}\right){\ⅇ}^{\frac{\left(5+\sqrt{17}\right)\ln \left(x\right)}{2}}}{2x}\\ \text{•}& & \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=1\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=1\right\}}=10\\ & & 10=-\frac{\mathrm{\_C1}\left(-5+\sqrt{17}\right)}{2}+\frac{\mathrm{\_C2}\left(5+\sqrt{17}\right)}{2}\\ \text{•}& & \text{Solve for}\mathrm{\_C1}\end{array}$