# Differential Calculus Problems With Solution

[mage lang="en|es|en" source="flickr"]differential calculus problems with solution[/mage]
Calculus problem. Help with separating differential equations?

Suppose dy/dθ = y^5cosθsin^5θ.

1. Separate the differential equation, then integrate both sides.
2. Write the general solution as a function y(θ).

Help is much appreciated. Thanks!

1)
dy / dθ = y^5 cosθ sin^5θ

dy = y^5 cosθ sin^5θ dθ

(1/y^5) dy = cosθ sin^5θ dθ

That is the separation.

2)
Let’s integrate ∫cosθ sin^5θ dθ = ∫sin^5θ cosθ dθ
You will have to use integration by parts.

u = sin^5θ
du = 5(sin^4θ)cosθ dθ

dv = cosθ dθ
v = sinθ

∫u dv = uv – ∫v du

∫sin^5θ cosθ dθ = (sin^5θ)(sin^θ) – ∫sinθ*5(sin^4θ)*cosθ dθ
∫(sin^5θ)(cosθ) dθ = (sin^6θ) – 5∫(sin^5θ)(cosθ) dθ

The left side and right side have a common integral so add 5∫(sin^5θ)(cosθ) dθ to both sides.

6∫(sin^5θ)(cosθ) dθ = (sin^6θ)

Divide both sides by 6.

Thus
∫(sin^5θ)(cosθ) dθ = [(sin^6θ)/6] + C

Let’s integrate
∫(1/y^5) dy = ∫(y^-5) dy
(y^-4)/-4

Finally:

[(y^-4)/-4] = [(sin^6θ)/6] + C

You can simply further more if you like.

Differential Calculus – Basic Derivation of Polynomials – Part 1

Category: Calculus Problems
You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
One Response
1. bryll says:

y=[(3y+4y^2)^2-3y