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Let f be the function satisfying f'(x)=x[tex]\sqrt{f(x)}[/tex] for all real numbers where f(3)=25.

a. Find f''(3)

b. Write an expression for y-f(x0 by solving the differential equation [tex]\frac{dy}{dx}[/tex] = x[tex]\sqrt{y}[/tex] with the initial condition of f(3)=25.

For a, I got [tex]\frac{x^2}{2}[/tex]+[tex]\sqrt{f(x)}[/tex], so my answer was [tex]\frac{19}{2}[/tex].

For b, I immediately substituted, getting dy/dx=3sqrt(25). then, dy/dx=15 -> dy=15dx -> integrate, y=15x+c, and since the initial condition is f(3)=25, by substitution, C=-20. My answer in the end was y=15x-20.

Am I on the right track? Thanks.

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# Homework Help: Differential Equation expression help

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