Internal problem ID [5202]
Internal file name [OUTPUT/4695_Sunday_June_05_2022_03_03_32_PM_35441377/index.tex]
Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill
2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems.
page 248
Problem number: Problem 24.19.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "linear", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_linear, `class A`]]
\[ \boxed {y^{\prime }+2 y={\mathrm e}^{x}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=2\\ q(x) &={\mathrm e}^{x} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+2 y = {\mathrm e}^{x} \end {align*}
The domain of \(p(x)=2\) is \[
\{-\infty
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode
and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )+2 Y \left (s \right ) = \frac {1}{s -1}\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} s Y \left (s \right )-1+2 Y \left (s \right ) = \frac {1}{s -1} \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {s}{\left (s -1\right ) \left (s +2\right )} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {2}{3 \left (s +2\right )}+\frac {1}{3 s -3} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2}{3 \left (s +2\right )}\right ) &= \frac {2 \,{\mathrm e}^{-2 x}}{3}\\ \mathcal {L}^{-1}\left (\frac {1}{3 s -3}\right ) &= \frac {{\mathrm e}^{x}}{3} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {2 \,{\mathrm e}^{-2 x}}{3}+\frac {{\mathrm e}^{x}}{3} \]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {2 \,{\mathrm e}^{-2 x}}{3}+\frac {{\mathrm e}^{x}}{3} \\
\end{align*} Verification of solutions
\[
y = \frac {2 \,{\mathrm e}^{-2 x}}{3}+\frac {{\mathrm e}^{x}}{3}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+2 y={\mathrm e}^{x}, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 y+{\mathrm e}^{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+2 y={\mathrm e}^{x} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+2 y\right )=\mu \left (x \right ) {\mathrm e}^{x} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+2 y\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=2 \mu \left (x \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )={\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int \mu \left (x \right ) {\mathrm e}^{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \mu \left (x \right ) {\mathrm e}^{x}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (x \right ) {\mathrm e}^{x}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )={\mathrm e}^{2 x} \\ {} & {} & y=\frac {\int {\mathrm e}^{2 x} {\mathrm e}^{x}d x +c_{1}}{{\mathrm e}^{2 x}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\frac {{\mathrm e}^{3 x}}{3}+c_{1}}{{\mathrm e}^{2 x}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{3 x}+3 c_{1} \right ) {\mathrm e}^{-2 x}}{3} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {1}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {2}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {2}{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{3} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.343 (sec). Leaf size: 15
\[
y \left (x \right ) = \frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{3}
\]
✓ Solution by Mathematica
Time used: 0.04 (sec). Leaf size: 21
\[
y(x)\to \frac {1}{3} e^{-2 x} \left (e^{3 x}+2\right )
\]
5.3.2 Solving as laplace ode
5.3.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
dsolve([diff(y(x),x)+2*y(x)=exp(x),y(0) = 1],y(x), singsol=all)
DSolve[{y'[x]+2*y[x]==Exp[x],{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]