Question

Find the general solution of each of the following differential equations. Compare a computer solution and, if necessary, reconcile it with yours. y+ytanhx=2exy^{\prime}+y \tanh x=2 e^{x}.

Solution

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Answered 1 year ago
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From eq.(3.4), we get

I=tanhx  dx=ln(coshx)eI=eln(coshx)=coshx\begin{align*} I &= \tanh x \; dx = \ln \left( \cosh x\right) \\ e^I &= e^{ \ln \left( \cosh x\right) } = \cosh x \end{align*}

Using eq.(3.9) we can find a solution for this differential equation

yeI=(2ex)coshx  dx=e2x2+x+Cy=e2x/2+x+Ccoshx\begin{align*} ye^I &= \int \left( 2e^x \right) \cosh x \; dx = \frac{e^{2x}}{2}+ x+C \\ y &= \frac{e^{2x}/2+x+C}{\cosh x} \end{align*}

Using the definition of the hyperbolic cosine

coshz=ez+ez2=1+e2z2ez=e2z+12ez\begin{align*} \cosh z &= \frac{e^z+e^{-z}}{2} = \frac{1+e^{-2z}}{2e^{-z}} = \frac{e^{2z}+1}{2e^z} \end{align*}

(actually, we car going to use the last one) we can rewrite the solution as follow

y=2ex(e2x/2+x+C)e2x+1=ex(e2x+2x+C1)e2x+1\begin{align*} y &= \frac{2e^x(e^{2x}/2+x+C)}{e^{2x}+1} = \frac{e^x(e^{2x}+2x+C_1)}{e^{2x}+1} \end{align*}

On Wolfram Mathematica 11.1, we can write DSolve[y’[x] + y [x] Tanh[x] == 2 Eˆx, y[x], x]\texttt{DSolve[y'[x] + y [x] Tanh[x] == 2 E\^{}x, y[x], x]}, which will give us

y=2ex(e2x/2+x)1+e2x+exC1+e2x=ex(e2x+2x+C1)e2x+1\begin{align*} y&= \frac{2e^x (e^{2x}/2+x)}{1+e^{2x}}+\frac{e^xC}{1+e^{2x}} = \frac{e^x(e^{2x}+2x+C_1)}{e^{2x}+1} \end{align*}

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