How do you find a function f(x), which, when multiplied by its derivative, gives you $x^{3}$ $x^3$ , and for which $f (0) = 4$ $f(0) = 4$ ?

Question

4034 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-01T15:46:35

The answer is: $y = \sqrt{\frac{x^{4}}{2} + 16}$ $y=sqrt(x^4/2+16)$ .

This is a separable variable first-order differential equation:

$yy'=x^3rArry(dy)/(dx)=x^3rArrydy=x^3dxrArr$

$intydy=intx^3dxrArry^2/2=x^4/4+crArry=+-sqrt(x^4/2+2c)$

Since the initial condition is:

$f(0)=4$ , we have to choose the positive one, and than we can find the constant $c$ :

$4=sqrt(0/2+2c)rArr16=2crArrc=8$ .

So the function is:

$y=sqrt(x^4/2+16)$ .

How do you find a function f(x), which, when multiplied by its derivative, gives you x^3x3x^3, and for which f(0) = 4f(0)=4f(0) = 4?