How do you find the solution of #(dy)/(dx) = -4x^3 e^(-x^4)#, #y(0) = 8#?

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Answer 1 · 2017-05-07T16:16:57

#y=e^(-x^4)+7#

Separating:

#dy=-4x^3e^(-x^4)dx#

#intdy=int-4x^3e^(-x^4)dx#

#y=int-4x^3e^(-x^4)dx#

On the right, use the substitution #u=-x^4#. Differentiating this shows that #du=-4x^3dx#. Both of these are in the integral:

#y=inte^(-x^4)(-4x^3dx)#

#y=inte^udu#

#y=e^u+C#

#y=e^(-x^4)+C#

We can use the condition #y(0)=8#:

#8=e^0+C#

#8=1+C#

#C=7#

So:

#y=e^(-x^4)+7#