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tester [92]
3 years ago
9

Solve this problem by using variation of parameters method. y''-y=coshx.

Mathematics
1 answer:
Gekata [30.6K]3 years ago
3 0
y''-y=0\implies r^2-1=0\implies r=\pm1
\implies y_c=C_1e^x+C_2e^{-x}={C^*}_1\underbrace{\cosh x}_{y_1}+{C^*}_2\underbrace{\sinh x}_{y_2}

For the nonhomogeneous ODE

y''-y=\underbrace{\cosh x}_{f(x)}

we're looking for a particular solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the two fundamental solutions.

We have

W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1

so we're left with

u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x
u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x

so that the particular solution is

y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x
y_p=-\dfrac12\cosh x+\dfrac12x\sinh x

As y_1 already accounts for the \cosh x term in y_p, we're left with the general solution

y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x
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