The maximum value of $f(x)$

Question

$f(x)$ is twice differentiable function such that $f(x)+f''(x)=-x\big|sinx\big|f'(x)$ where $x\geq0.$Given that $f(0)=-3,f'(0)=4$ the maximum value of $f(x)$ is$?$

My Try:To say, I was never taught these second order DE's but this question appeared under the topic maxima-minima,however I am familiar with first order DE's.Any particular way to solve these type of questions(I mean second order DE's)$?$

Answer 1

$$ff'+f'f''=-x|\sin x|(f'(x))^2$$ $$d(f^2+(f')^2) \leq 0$$ can you continue after this

Answer 2

Maxima occur when $f'(x) = 0$, $f'(x)$ fails to exist, or at endpoints.

The domain has one endpoint, at $0$, but because $f'(0) = 4$, we can conclude that it is a local minimum and definitely not the maximum.

Since they say $f(x)$ is twice differentiable, $f'(x)$ must exist everywhere.

That leaves the final option, $f'(x) = 0$. Plugging that in, the right hand side disappears and you are left with a much simpler equation. However, as @LutzLehmann pointed out, that equation only holds at the maxima themselves.

If it were not for that factor of $x$ I would pursue showing that $f(x)$ must be periodic. Since it is there, that approach fails.

General ideas: Look for traits the function must possess. Is it always increasing? (Then it would not attain a maximum, though it might approach one.) $f''(0) = 3$ when you plug in the initial conditions. Try restrictions. Suppose $f''(x)$ is always positive? Bounded? Can you factor the unknown function? Write $f(x) = g(x)\cdot e^x$, or other trial functions. And of course, there is always numerical approximation as the fallback. You can get a rough feel for the look of the equation via Euler's method.

Derivatives are a science, integrals are an art, and differential equations are a black art. Good luck.

Answer 3

To have any solution at all, consider the numerical solution. Use an event mechanism to find the roots of the derivative. Here in python, one could do almost literally the same in matlab/octave

def ode(x,f): return [f[1], -f[0]-x*abs(np.sin(x))*f[1]]
def evt(x,f): return f[1]
evt.direction=-1

sol = solve_ivp(ode,[0,10],[-3,4], events=evt, atol=1e-8, rtol=1e-9)

print(sol.t_events[0])
print(sol.y_events[0][:,0])

This gives exactly one event, there are no other extrema

x    = [1.8246435]
f(x) = [2.7319696]

The plot makes it clear that in the second $\pi$ interval the friction coefficient $x|\sin x|$ becomes so large that the oscillator is over-dampened, the solution falls directly to the asymptote $y=0$.