Reduce this equation to first order: $y''+y=\frac12\cosh(y)$

Question

Equation is

$$y''+y=\frac{1}{2}\cosh(y) $$ $ y(0)=0$, $y'(0)=\sqrt{0.3}$

What is the max value of y?

Need help for my revision for exam. thank you.

Answer 1

Multiply with $2y'$ and integrate once to get $$ y'^2+y^2=\sinh(y)+C $$ where you find from initial conditions that $C=0.3$.

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The maximal value of $y$ is taken at one of the places where $y'=0$ so that you have to solve $$ y^2=\sinh(y)+0.3 $$

In the graph one can see that there are roots close to $-0.25$, $1.8$ and $2.4$. The solution of the ODE however can only move below the zero level as $y'^2$ is non-negative. Thus the max of the solution to the given IVP will have its maximum at $1.8$

Answer 2

Hint:

Let $s=y’$.

This yields: $$s\frac{ds}{dy}=\frac1{2}\cosh y-y$$

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$$sds=\left(\frac1{2}\cosh y-y\right)dy$$

$$C_1+\frac{s^2}2=-\frac{y^2}2+\frac{\sinh y}2$$

Clearly, $C_1=-0.15$.

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The max value $y$ can attain, should occur at $s=0$.

Thus, the maximum value of $y$, call it $M$, satisfies

$$\frac{\sinh M-M^2}2=-0.15$$

and $$y’’=\frac{\cosh M}2-M<0$$

Numerically, $M\approx 1.8$.