First Order IVPs - Maple Help

ODE Steps for First Order IVPs

Overview

 • This help page gives a few examples of using the command ODESteps to solve first order initial value problems.
 • See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.

Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
 > $\mathrm{ivp1}≔\left\{{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\mathrm{diff}\left(z\left(t\right),t\right)=0,z\left(3\right)=1\right\}$
 ${\mathrm{ivp1}}{≔}\left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}$ (1)
 > $\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}{=}{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\\ {}& {}& {\int }\frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{\int }{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}z{}\left(3\right)=1\\ {}& {}& {-}\frac{{1}}{{2}}{+}{\mathrm{ln}}{}\left({2}\right){=}{-}\frac{{15}}{{2}}{-}{\mathrm{ln}}{}\left({2}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=7+2{}\mathrm{ln}{}\left(2\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\end{array}$ (2)
 > $\mathrm{ivp2}≔\left\{2xy\left(x\right)-9{x}^{2}+\left(2y\left(x\right)+{x}^{2}+1\right)\mathrm{diff}\left(y\left(x\right),x\right)=0,y\left(0\right)=1\right\}$
 ${\mathrm{ivp2}}{≔}\left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (3)
 > $\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{▫}& {}& \text{Check if ODE is exact}\\ {}& \text{◦}& \text{ODE is exact if the lhs is the total derivative of a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{C}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{function}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Compute derivative of lhs}\\ {}& {}& \frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){+}\left(\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Evaluate derivatives}\\ {}& {}& {2}{}{x}{=}{2}{}{x}\\ {}& \text{◦}& \text{Condition met, ODE is exact}\\ \text{•}& {}& \text{Exact ODE implies solution will be of this form}\\ {}& {}& \left[{F}{}\left({x}{,}{y}\right){=}{\mathrm{C1}}{,}{M}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){,}{N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right]\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{by integrating}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}M{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{\int }\left({-}{9}{}{{x}}^{{2}}{+}{2}{}{x}{}{y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Take derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\\ \text{•}& {}& \text{Compute derivative}\\ {}& {}& {{x}}^{{2}}{+}{2}{}{y}{+}{1}{=}{{x}}^{{2}}{+}\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Isolate for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆy}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right){=}{2}{}{y}{+}{1}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& {\mathrm{_F1}}{}\left({y}\right){=}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into the solution of the ODE}\\ {}& {}& {-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}{=}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& \left\{{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}{,}{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right\}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solution does not satisfy initial condition}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{2}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\end{array}$ (4)
 > $\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right),x\right)-y\left(x\right)-x\mathrm{exp}\left(x\right)=0,y\left(a\right)=b\right\}$
 ${\mathrm{ivp3}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}$ (5)
 > $\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$
 $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate the derivative}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{y}{}\left({x}\right){+}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{The ODE is linear; multiply by an integrating factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Assume the lhs of the ODE is the total derivative}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left(\mathrm{\mu }{}\left(x\right){}y{}\left(x\right)\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right)\right){}{y}{}\left({x}\right){+}{\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\\ \text{•}& {}& \text{Isolate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right){=}{-}{\mathrm{\mu }}{}\left({x}\right)\\ \text{•}& {}& \text{Solve to find the integrating factor}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){=}{{ⅇ}}^{{-}{x}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right)\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate the integral on the lhs}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right){=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{\mathrm{\mu }}{}\left({x}\right)}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)={ⅇ}^{-x}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{{ⅇ}}^{{-}{x}}{}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Evaluate the integrals on the rhs}\\ {}& {}& {y}{}\left({x}\right){=}\frac{\frac{{{x}}^{{2}}}{{2}}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(a\right)=b\\ {}& {}& {b}{=}\frac{{{ⅇ}}^{{a}}{}\left({{a}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{-}\frac{{{ⅇ}}^{{a}}{}{{a}}^{{2}}{-}{2}{}{b}}{{2}{}{{ⅇ}}^{{a}}}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=-\frac{{ⅇ}^{a}{}{a}^{2}-2{}b}{2{}{ⅇ}^{a}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\end{array}$ (6)