Integration: Let f be the function f'(x) = xsqrt[f(x)]....

[paulxzt]

Let f be the function f'(x) = xsqrt[f(x)] for all real numbers x where f(3) = 25.

a) find f''(3)

b) write an expression y = f(x) by solving the differential equation dy/dx = x*sqrt[y] with the initial condition f(3) = 25.


To solve for a would I just take the derivative of f'(x)? If so, I am confused about what I need to do since there is an f(x) in the equation.

For b do I integrate it by separating x and y?

integral of dy/sqrt[y] = integral of x(dx)

y = .278e^1/2x^2 ? Can someone check please.

Thank you

 

[skeeter]

\(\displaystyle \L f'(x) = x[f(x)]^{\frac{1}{2}}\)

product rule + chain rule ...

\(\displaystyle \L f''(x) = x*\frac{1}{2}[f(x)]^{-\frac{1}{2}}*f'(x) + [f(x)]^{\frac{1}{2}}\)

\(\displaystyle \L f(3) = 25\)

\(\displaystyle \L f'(3) = 3[f(3)]^{\frac{1}{2}} = 15\)

\(\displaystyle \L f''(3) = 3*\frac{1}{2}[f(3)]^{-\frac{1}{2}}*f'(3) + [f(3)]^{\frac{1}{2}}\)

I'll let you finish the evaluation of f''(3).

------------------------------------------------

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

separate variables ...

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

integrate ...

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

use the initial condition to find the constant of integration, then solve for y.

you should get \(\displaystyle \L y = \frac{(x^2+1)^2}{4}\)

 

[paulxzt]

\(\displaystyle \L f'(x) = x[f(x)]^{\frac{1}{2}}\)



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

separate variables ...

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

integrate ...

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

use the initial condition to find the constant of integration, then solve for y.

you should get \(\displaystyle \L y = \frac{(x^2+1)^2}{4}\)

The part you integrated, shouldn't it be ln sqrt[y] ?

 

[skeeter]

no.

\(\displaystyle \L \frac{1}{\sqrt{y}} = y^{-\frac{1}{2}}\)

the antiderivative of \(\displaystyle \L y^{-\frac{1}{2}}\) is \(\displaystyle \L 2y^{\frac{1}{2}}\) ... check it by taking the derivative.

 

[paulxzt]

no.

\(\displaystyle \L \frac{1}{\sqrt{y}} = y^{-\frac{1}{2}}\)

the antiderivative of \(\displaystyle \L y^{-\frac{1}{2}}\) is \(\displaystyle \L 2y^{\frac{1}{2}}\) ... check it by taking the derivative.

oh. what about the x^2 +c ? why isn't it 1/2x^2 + c ? taking the derivative of (1/2)x^2 is x ?

 

[skeeter]

yes ... mea culpa, I left of the divisor of 2.

correction ...

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

initial condition y(3) = 25

\(\displaystyle \L 2\sqrt{25} = \frac{9}{2} + C\) ... \(\displaystyle \L C = \frac{11}{2}\)

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

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

\(\displaystyle \L y = \frac{(x^2 + 11)^2}{16}\)

better?