# Differential Equation

• Feb 15th 2010, 05:16 AM
NeilT
Differential Equation
Hi there I've been having a problem with this differential equation. I have the answer y = c(sin x)^2 for part (a) but it doesn't seem right. I'd appreciate any help with ether part.

(a) Given the differential equation

(sin x)(dy/dx) - 2ycos x = 0,

find the general solution, expressing y explicitly in terms of x.

(b) Find the general solution of

(sin x)(dy/dx) - 2ycos x = 3(sin x)^3.
• Feb 15th 2010, 05:33 AM
vince
Quote:

Originally Posted by NeilT
Hi there I've been having a problem with this differential equation. I have the answer y = c(sin x)^2 for part (a) but it doesn't seem right. I'd appreciate any help with ether part.

(a) Given the differential equation

(sin x)(dy/dx) - 2ycos x = 0,

find the general solution, expressing y explicitly in terms of x.

(b) Find the general solution of

(sin x)(dy/dx) - 2ycos x = 3(sin x)^3.

Well for part(a),

you can separate the variables to get:
$\displaystyle \frac{dy}{dx}\frac{1}{2y}=\frac{cosx}{sinx}$
=

$\displaystyle \frac{dy}{2y}=\frac{cosxdx}{sinx}$

Now take the integral for each side. To do so, substitute for the quantity in the denominator. So for the right side, that'd be $\displaystyle u=sinx$, which would mean $\displaystyle du=cosxdx$ so that the right side integral evaluates to: $\displaystyle ln(sinx)+C$
Doing the same for the left gets you $\displaystyle \frac{lny}{2}$.
Settitng these equal to each other and solving for y leads to

$\displaystyle lny = 2ln(sinx)+C$
=
$\displaystyle exp(lny) = exp(2ln(sinx)+C)$
=
$\displaystyle y = Csin^2(x)$

I hope that was done in enough detail. Now try the second one. The key was to try to separate the variables first.
• Feb 15th 2010, 05:43 AM
NeilT
Thanks Vince I'm gald to see you agree with me for (a) even though I did it slightly differently:

dividing by $\displaystyle sin x$ gives

[tex]\frac(dy,dx)[tex]
• Feb 15th 2010, 05:52 AM
vince
Quote:

Originally Posted by NeilT
Thanks Vince I'm gald to see you agree with me for (a) even though I did it slightly differently:

dividing by $\displaystyle sin x$ gives

[tex]\frac(dy,dx)[tex]

oh good. I thought your answer was from an answer book which was i why i showed all the steps. so since you have a grasp of what is required, why is part(b) any more difficult? go for it lad.
• Feb 15th 2010, 05:56 AM
NeilT
Thanks for ths Vince my biggest problem is actually rememberng how to use LATEX - it's been nearly 4 years since I last used it.

I have an answer for (b) and would be interested to see if you agree:

$\displaystyle y = 3x(sinx)^2 + c(sinx)^2$
• Feb 15th 2010, 05:59 AM
NeilT
For info I used an Integrating Factor of $\displaystyle \frac{1}{sin^2 x}$

Which seemed to make things much easier
• Feb 15th 2010, 06:31 AM
vince
You are correct in that i agree. One can always check by taking the solution and substituting it into the differential. The solution by definition satisfies the differential relation. Well done.
• Feb 15th 2010, 06:33 AM
NeilT
Quote:

Originally Posted by vince
One can always check by taking the solution and substituting it into the differential.

Of course you can! I'll remember to do that in future. Thanks for your help!