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