2.5.17 Problem 17

Internal problem ID [8977]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 17
Date solved : Friday, April 25, 2025 at 05:32:35 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+3y^{\prime }-4y& =6{\mathrm{e}}^{2t-2}\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(1\right)& =4\\ y^{\prime }\left(1\right)& =5\end{array}$$

Since both initial conditions are not at zero, then let

$$\begin{array}{rl}y(0)& =c_1\\ y^{\prime }(0)& =c_2\end{array}$$

Solving using the Laplace transform method. Let

$$\begin{array}{r}\mathcal{L}\left(y\right)=Y(s)\end{array}$$

Taking the Laplace transform of the ode and using the relations that

$$\begin{array}{rl}\mathcal{L}\left(y^{\prime }\right)& =sY(s)-y\left(0\right)\\ \mathcal{L}\left(y^{\prime \prime }\right)& =s^{2}Y(s)-y^{\prime }(0)-sy\left(0\right)\end{array}$$

The given ode now becomes an algebraic equation in the Laplace domain.

$$\begin{array}{}\text{(1)}& s^{2}Y\left(s\right)-y^{\prime }\left(0\right)-sy\left(0\right)+3sY\left(s\right)-3y\left(0\right)-4Y\left(s\right)=\frac{6{\mathrm{e}}^{-2}}{s-2}\end{array}$$

Substituting the initial conditions in above in Eq (1) gives

$$s^{2}Y\left(s\right)-c_2-s{c}_1+3sY\left(s\right)-3c_1-4Y\left(s\right)=\frac{6{\mathrm{e}}^{-2}}{s-2}$$

Solving the above equation for $Y(s)$ results in

$$Y(s)=\frac{s^2{c}_1+s{c}_1+c_{2}s+6{\mathrm{e}}^{-2}-6c_1-2c_2}{(s-2)(s^2+3s-4)}$$

Applying partial fractions decomposition results in

$$Y(s)=\frac{{\mathrm{e}}^{-2}}{s-2}+\frac{\frac{4c_1}{5}+\frac{c_2}{5}-\frac{6{\mathrm{e}}^{-2}}{5}}{s-1}+\frac{\frac{c_1}{5}-\frac{c_2}{5}+\frac{{\mathrm{e}}^{-2}}{5}}{s+4}$$

The inverse Laplace of each term above is now found, which gives

$$\begin{array}{rl}{\mathcal{L}}^{-1}\left(\frac{{\mathrm{e}}^{-2}}{s-2}\right)& ={\mathrm{e}}^{2t-2}\\ {\mathcal{L}}^{-1}\left(\frac{\frac{4c_1}{5}+\frac{c_2}{5}-\frac{6{\mathrm{e}}^{-2}}{5}}{s-1}\right)& =\frac{{\mathrm{e}}^t(4c_1+c_2-6{\mathrm{e}}^{-2})}{5}\\ {\mathcal{L}}^{-1}\left(\frac{\frac{c_1}{5}-\frac{c_2}{5}+\frac{{\mathrm{e}}^{-2}}{5}}{s+4}\right)& =\frac{(c_1-c_2+{\mathrm{e}}^{-2}){\mathrm{e}}^{-4t}}{5}\end{array}$$

Adding the above results and simplifying gives

$$y={\mathrm{e}}^{2t-2}+\frac{{\mathrm{e}}^t(4c_1+c_2-6{\mathrm{e}}^{-2})}{5}+\frac{(c_1-c_2+{\mathrm{e}}^{-2}){\mathrm{e}}^{-4t}}{5}$$

Solving for $c_1$ and $c_2$ from the given initial conditions results in

$$y={\mathrm{e}}^{2t-2}+3{\mathrm{e}}^t{\mathrm{e}}^{-1}$$

Simplifying the solution gives

$$y=3{\mathrm{e}}^{-1+t}+{\mathrm{e}}^{2t-2}$$

Figure 2.194: Solution plot

✓ Maple. Time used: 0.106 (sec). Leaf size: 17

ode:=diff(diff(y(t),t),t)+3*diff(y(t),t)-4*y(t) = 6*exp(2*t-2); 
ic:=y(1) = 4, D(y)(1) = 5; 
dsolve([ode,ic],y(t),method='laplace');
 

$$y={\mathrm{e}}^{2t-2}+3{\mathrm{e}}^{t-1}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [\frac{d}{dt}\frac{d}{dt}y\left(t\right)+3\frac{d}{dt}y\left(t\right)-4y\left(t\right)=6{\mathrm{e}}^{2t-2},y\left(1\right)=4,\left(\frac{d}{dt}y\left(t\right)\right)|{}_{\left\{t=1\right\}}=5]\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2+3r-4=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & (r+4)(r-1)=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-4,1)\\ \bullet & & \text{1st solution of the homogeneous ODE}\\ & & y_1\left(t\right)={\mathrm{e}}^{-4t}\\ \bullet & & \text{2nd solution of the homogeneous ODE}\\ & & y_2\left(t\right)={\mathrm{e}}^t\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{y}_1\left(t\right)+\mathit{C}\mathit{2}{y}_2\left(t\right)+y_p\left(t\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-4t}+\mathit{C}\mathit{2}{\mathrm{e}}^t+y_p\left(t\right)\\ ◻& & \text{Find a particular solution}{y}_p\left(t\right)\text{of the ODE}\\ & \circ & \text{Use variation of parameters to find}{y}_p\text{here}f\left(t\right)\text{is the forcing function}\\ & & [y_p\left(t\right)=-y_1\left(t\right)\int \frac{y_2\left(t\right)f\left(t\right)}{W(y_1\left(t\right),y_2\left(t\right))}dt+y_2\left(t\right)\int \frac{y_1\left(t\right)f\left(t\right)}{W(y_1\left(t\right),y_2\left(t\right))}dt,f\left(t\right)=6{\mathrm{e}}^{2t-2}]\\ & \circ & \text{Wronskian of solutions of the homogeneous equation}\\ & & W(y_1\left(t\right),y_2\left(t\right))=\left[\begin{array}{cc}{\mathrm{e}}^{-4t}& {\mathrm{e}}^t\\ -4{\mathrm{e}}^{-4t}& {\mathrm{e}}^t\end{array}\right]\\ & \circ & \text{Compute Wronskian}\\ & & W(y_1\left(t\right),y_2\left(t\right))=5{\mathrm{e}}^{-3t}\\ & \circ & \text{Substitute functions into equation for}{y}_p\left(t\right)\\ & & y_p\left(t\right)=-\frac{6(-\int {\mathrm{e}}^{-2+t}dt{\mathrm{e}}^{5t}+\int {\mathrm{e}}^{-2+6t}dt){\mathrm{e}}^{-4t}}{5}\\ & \circ & \text{Compute integrals}\\ & & y_p\left(t\right)={\mathrm{e}}^{2t-2}\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-4t}+\mathit{C}\mathit{2}{\mathrm{e}}^t+{\mathrm{e}}^{2t-2}\\ ◻& & \text{Check validity of solution}y\left(t\right)=\mathit{\text{\_C1}}{\mathrm{e}}^{-4t}+\mathit{\text{\_C2}}{\mathrm{e}}^t+{\mathrm{e}}^{2t-2}\\ & \circ & \text{Use initial condition}y\left(1\right)=4\\ & & 4=\mathit{\text{\_C1}}{\mathrm{e}}^{-4}+\mathit{\text{\_C2}}\mathrm{e}+1\\ & \circ & \text{Compute derivative of the solution}\\ & & \frac{d}{dt}y\left(t\right)=-4\mathit{\text{\_C1}}{\mathrm{e}}^{-4t}+\mathit{\text{\_C2}}{\mathrm{e}}^t+2{\mathrm{e}}^{2t-2}\\ & \circ & \text{Use the initial condition}\left(\frac{d}{dt}y\left(t\right)\right)|{}_{\left\{t=1\right\}}=5\\ & & 5=-4\mathit{\text{\_C1}}{\mathrm{e}}^{-4}+\mathit{\text{\_C2}}\mathrm{e}+2\\ & \circ & \text{Solve for}\mathit{\text{\_C1}}\text{and}\mathit{\text{\_C2}}\\ & & \{\mathit{\text{\_C1}}=0,\mathit{\text{\_C2}}=\frac{3}{\mathrm{e}}\}\\ & \circ & \text{Substitute constant values into general solution and simplify}\\ & & y\left(t\right)=3{\mathrm{e}}^{t-1}+{\mathrm{e}}^{2t-2}\\ \bullet & & \text{Solution to the IVP}\\ & & y\left(t\right)=3{\mathrm{e}}^{t-1}+{\mathrm{e}}^{2t-2}\end{array}$$

✓ Mathematica. Time used: 0.076 (sec). Leaf size: 18

ode=D[y[t],{t,2}]+3*D[y[t],t]-4*y[t]==6*Exp[2*t-2]; 
ic={y[1]==4,Derivative[1][y][1]==5}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to e^{t-2}(e^t+3e)$$

✓ Sympy. Time used: 0.240 (sec). Leaf size: 17

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-4*y(t) - 6*exp(2*t - 2) + 3*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0) 
ics = {y(1): 4, Subs(Derivative(y(t), t), t, 1): 5} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=\frac{3e^t}{e}+e^{2t-2}$$