The ByLaplaceTransform(ODE, IC, y(x)) command finds the solution of a 2nd order linear ordinary differential equation ODE with initial conditions IC by using the Laplace transform.

The coefficients of y(x) and its derivatives must be constant.

Use the option output=steps to make this command return an annotated step-by-step solution.  Further control over the format and display of the step-by-step solution is available using the options described in Student:-Basics:-OutputStepsRecord.  The options supported by that command can be passed to this one.

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{ODEs}\right]\left[\mathrm{Solve}\right]\right)\:$

$\mathrm{ode1}≔\mathrm{diff}\left(x\left(t\right)\,t\,t\right)+2\mathrm{diff}\left(x\left(t\right)\,t\right)+2x\left(t\right)=0$

$\mathrm{ode1}≔\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)+2\frac{\ⅆ}{\ⅆt}x\left(t\right)+2x\left(t\right)=0$

$\mathrm{ic1}≔\left\{\mathrm{eval}\left(\mathrm{diff}\left(x\left(t\right)\,t\right)\,t=0\right)=-2\,x\left(0\right)=1\right\}$

$\mathrm{ic1}≔\left\{\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}{\phantom{\left\{t=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}}{\left\{t=0\right\}}=\mathrm{-2}\,x\left(0\right)=1\right\}$

$\mathrm{sol1}≔\mathrm{ByLaplaceTransform}\left(\mathrm{ode1}\,\mathrm{ic1}\,x\left(t\right)\right)$

$\mathrm{sol1}≔x\left(t\right)={\ⅇ}^{-t}\cos \left(t\right)-{\ⅇ}^{-t}\sin \left(t\right)$

$\mathrm{ode2}≔\mathrm{diff}\left(x\left(t\right)\,t\,t\right)+\mathrm{diff}\left(x\left(t\right)\,t\right)-6x\left(t\right)=\sin \left(t\right)$

$\mathrm{ode2}≔\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)+\frac{\ⅆ}{\ⅆt}x\left(t\right)-6x\left(t\right)=\sin \left(t\right)$

$\mathrm{ic2}≔\left\{\mathrm{eval}\left(\mathrm{diff}\left(x\left(t\right)\,t\right)\,t=0\right)=2\,x\left(0\right)=-3\right\}$

$\mathrm{ic2}≔\left\{\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}{\phantom{\left\{t=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}}{\left\{t=0\right\}}=2\,x\left(0\right)=\mathrm{-3}\right\}$

$\mathrm{sol2}≔\mathrm{ByLaplaceTransform}\left(\mathrm{ode2}\,\mathrm{ic2}\,x\left(t\right)\right)$

$\mathrm{sol2}≔x\left(t\right)=-\frac{34{\ⅇ}^{2t}}{25}-\frac{81{\ⅇ}^{-3t}}{50}-\frac{\cos \left(t\right)}{50}-\frac{7\sin \left(t\right)}{50}$

$\mathrm{ode3}≔\mathrm{diff}\left(x\left(t\right)\,t\,t\right)+4\mathrm{diff}\left(x\left(t\right)\,t\right)+4x\left(t\right)=\exp \left(2t\right)$

$\mathrm{ode3}≔\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)+4\frac{\ⅆ}{\ⅆt}x\left(t\right)+4x\left(t\right)={\ⅇ}^{2t}$

$\mathrm{ic3}≔\left\{\mathrm{eval}\left(\mathrm{diff}\left(x\left(t\right)\,t\right)\,t=0\right)=-2\,x\left(0\right)=1\right\}$

$\mathrm{ic3}≔\left\{\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}{\phantom{\left\{t=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}}{\left\{t=0\right\}}=\mathrm{-2}\,x\left(0\right)=1\right\}$

$\mathrm{sol3}≔\mathrm{ByLaplaceTransform}\left(\mathrm{ode3}\,\mathrm{ic3}\,x\left(t\right)\right)$

$\mathrm{sol3}≔x\left(t\right)=\frac{\left({\ⅇ}^{4t}-4t+15\right){\ⅇ}^{-2t}}{16}$

$\mathrm{ode4}≔\mathrm{diff}\left(x\left(t\right)\,t\,t\right)-6\mathrm{diff}\left(x\left(t\right)\,t\right)+13x\left(t\right)=t$

$\mathrm{ode4}≔\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)-6\frac{\ⅆ}{\ⅆt}x\left(t\right)+13x\left(t\right)=t$

$\mathrm{ic4}≔\left\{\mathrm{eval}\left(\mathrm{diff}\left(x\left(t\right)\,t\right)\,t=0\right)=-2\,x\left(0\right)=1\right\}$

$\mathrm{ic4}≔\left\{\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}{\phantom{\left\{t=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}}{\left\{t=0\right\}}=\mathrm{-2}\,x\left(0\right)=1\right\}$

$\mathrm{sol4}≔\mathrm{ByLaplaceTransform}\left(\mathrm{ode4}\,\mathrm{ic4}\,x\left(t\right)\right)$

$\mathrm{sol4}≔x\left(t\right)=\frac{6}{169}+\frac{t}{13}+\frac{\left(163\cos \left(2t\right)-420\sin \left(2t\right)\right){\ⅇ}^{3t}}{169}$

$\mathrm{ode5}≔\mathrm{diff}\left(x\left(t\right)\,t\,t\right)+4\mathrm{diff}\left(x\left(t\right)\,t\right)+4x\left(t\right)=\exp \left(-2t\right)$

$\mathrm{ode5}≔\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)+4\frac{\ⅆ}{\ⅆt}x\left(t\right)+4x\left(t\right)={\ⅇ}^{-2t}$

$\mathrm{ic5}≔\left\{\mathrm{eval}\left(\mathrm{diff}\left(x\left(t\right)\,t\right)\,t=0\right)=-2\,x\left(0\right)=2\right\}$

$\mathrm{ic5}≔\left\{\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}{\phantom{\left\{t=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}}{\left\{t=0\right\}}=\mathrm{-2}\,x\left(0\right)=2\right\}$

$\mathrm{sol5}≔\mathrm{ByLaplaceTransform}\left(\mathrm{ode5}\,\mathrm{ic5}\,x\left(t\right)\right)$

$\mathrm{sol5}≔x\left(t\right)=\frac{{\ⅇ}^{-2t}{\left(t+2\right)}^2}{2}$

The Student[ODEs][Solve][ByLaplaceTransform] command was introduced in Maple 2021.

For more information on Maple 2021 changes, see Updates in Maple 2021.

The Student[ODEs][Solve][ByLaplaceTransform] command was updated in Maple 2022.

The output option was introduced in Maple 2022.

For more information on Maple 2022 changes, see Updates in Maple 2022.