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

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Answer 1 · 2015-03-01T15:46:35

The answer is: #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)#.