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

Dec 17, 2016

$h \left(x\right) = 2 {x}^{4} + 5 x - 3$.

#### Explanation:

Start by writing in Lebeniz notation.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 {x}^{3} + 5$

This is a separable differential equation, so multiply both sides by $\mathrm{dx}$.

$\mathrm{dy} = 8 {x}^{3} + 5 \mathrm{dx}$

Integrate both sides.

$\int \left(\mathrm{dy}\right) = \int \left(8 {x}^{3} + 5\right) \mathrm{dx}$

Use $\int \left({x}^{n}\right) \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1} + C$ to solve the right hand side.

$y = 2 {x}^{4} + 5 x + C$

$h \left(x\right) = 2 {x}^{4} + 5 x + C$

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

$4 = 2 {\left(1\right)}^{4} + 5 \left(1\right) + C$

$4 = 2 + 5 + C$

$- 3 = C$

Then the solution to the differential equation is $h \left(x\right) = 2 {x}^{4} + 5 x - 3$.

Hopefully this helps!