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

Start by writing in Lebeniz notation.

Integrate both sides.

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To solve the differential equation ( h'(x) = 8x^3 + 5 ) with the initial condition ( h(1) = -4 ), you can integrate both sides of the equation with respect to ( x ).

[ \int h'(x) , dx = \int (8x^3 + 5) , dx ]

This gives:

[ h(x) = \int (8x^3 + 5) , dx ]

You then integrate each term separately:

[ h(x) = \int 8x^3 , dx + \int 5 , dx ]

[ h(x) = 2x^4 + 5x + C ]

To find the constant of integration ( C ), use the initial condition ( h(1) = -4 ):

[ -4 = 2(1)^4 + 5(1) + C ]

[ -4 = 2 + 5 + C ]

[ C = -4 - 2 - 5 ]

[ C = -11 ]

So, the solution to the differential equation is:

[ h(x) = 2x^4 + 5x - 11 ]

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