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

Question

Isabella Ackerman · Accepted Answer

#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!

Emily Ammerman · Answer

undefined 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:

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