How do you solve the differential equation given #h'(x)=8x^3+5#, h(1)=-4?

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Answer 1 · 2016-12-17T14:48:15

#h(x) = 2x^4 + 5x - 3#.

Start by writing in Lebeniz notation.

#dy/dx = 8x^3 + 5#

This is a separable differential equation, so multiply both sides by #dx#.

#dy = 8x^3 + 5 dx#

Integrate both sides.

#int(dy) = int(8x^3 + 5)dx#

Use #int(x^n)dx = (x^(n + 1))/(n + 1) + C# to solve the right hand side.

#y = 2x^4 + 5x + C#

#h(x) = 2x^4 + 5x + C#

Now, we know an input value (x) and an output value (y), so we can solve for #C#.

#4 = 2(1)^4 + 5(1) + C#

#4 = 2 + 5 + C#

#-3 = C#

Then the solution to the differential equation is #h(x) = 2x^4 + 5x - 3#.

Hopefully this helps!