Second derivative and differential equation

[Mooch22]

Let f be the function f'(x) = x(sq rt(f(x))) for all real numbers x, where f(3) = 25.

Find f ''(3).

Write an equation for y=f(x) by solving the differential equation (dy/dx) = x(sq. rt.(y)) with the initial condition f(3)=25.

How do you start these? Any help is greatly needed and appreciated!

 

[soroban]

Hello, Mooch22!

I'll do the first part now . . .

Let $\displaystyle f$ be the function: $\displaystyle \,f'(x)\:=\:x\cdot\sqrt{f(x))}$, where $\displaystyle f(3)\,=\,25$.

Find $\displaystyle \,f''(3)$

We have: \(\displaystyle \L\,f'(x)\;=\;x\cdot\left(f(x)\right)^{\frac{1}{2}}\)


Differentiate . . . (Product Rule and Chain Rule)

\(\displaystyle \L\;\;f''(x)\;=\;x\cdot\frac{1}{2}\left(f(x)\right)^{-\frac{1}{2}}\cdot f'(x)\,+\,1\cdot\left(f(x)\right)^{\frac{1}{2}}\)

We have: \(\displaystyle \L\,f''(x)\;=\;\frac{x\cdot f'(x)}{2\sqrt{f(x)}}\,+\,\sqrt{f(x)}\)


When $\displaystyle x\,=\,3$. we have: \(\displaystyle \L\,f''(3)\;=\;\frac{3\cdot f'(3)}{2\sqrt{f(3)}}\,+\,\sqrt{f(3)}\)

$\displaystyle \;\;$We are given: $\displaystyle \,f(3)\,=\,25$ . . . hence: $\displaystyle \,\sqrt{f(3)} \:=\:\sqrt{25}\:=\:5$

$\displaystyle \;\;$Since $\displaystyle \,f'(x)\:=\:x\sqrt{f(x)}$, then: $\displaystyle \,f'(3)\:=\:3\cdot\sqrt{f(3)}\:=\:3\cdot\sqrt{25}\:=\:15$


Therefore: \(\displaystyle \L\,f''(3)\;=\;\frac{3\cdot15}{2\cdot5}\,+\,5\;=\;\frac{19}{2}\)
.

 

[galactus]

You could separate variables.

They gave you:

\(\displaystyle \L\\\frac{dy}{dx}=x\sqrt{y}\)

\(\displaystyle \L\\\frac{dy}{\sqrt{y}}=xdx\)

Integrate:

\(\displaystyle \L\\\int{\frac{dy}{\sqrt{y}}}=\int{x}dx\)

\(\displaystyle \L\\2\sqrt{y}=\frac{x^{2}}{2}+C\)

Solve for y:

\(\displaystyle \L\\y=\frac{(x^{2}+C)^{2}}{16}\)

Using x=3 and y=25 and solving for C:

We find C=11 or -29

We use 11:

\(\displaystyle \L\\y'=\frac{x(x^{2}+11)}{4}\)

You can check this by subbing it back into your original equation.

Now, take the derivative of the derivative and sub in x=3:

\(\displaystyle \L\\y''=\frac{3x^{2}}{4}+\frac{11}{4}\)

Sub in x=3 and get 19/2