Differential equations - seperation of variables

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• Oct 15th 2009, 08:46 PM
wyasser
Differential equations - seperation of variables
Solve the given differential equation by Separation of variables.

(1+x^4)dy + x(1+4y^2)dx = 0, y(1)=0
(3x^2 + 9xy + 5y^2)dx – (6x^2 +4xy)dy = 0, y(2) = -6
(3y^2-x^2 / y^5)dy/dx + x/2y^4 = 0, y(1) = 1
Dr/dQ + rsecQ = cosQ
• Oct 16th 2009, 01:01 AM
craig
Quote:

Originally Posted by wyasser
Solve the given differential equation by Separation of variables.

(1+x^4)dy + x(1+4y^2)dx = 0, y(1)=0
(3x^2 + 9xy + 5y^2)dx – (6x^2 +4xy)dy = 0, y(2) = -6
(3y^2-x^2 / y^5)dy/dx + x/2y^4 = 0, y(1) = 1
Dr/dQ + rsecQ = cosQ

It will help people to answer your question if you write out your equations using LATEX.

For the first one:

$\displaystyle (1+x^4)dy + x(1+4y^2)dx = 0$

Here divide both sides by $\displaystyle (1+x^4)(1+4y^2)$, this will leave you with:

$\displaystyle \int\frac{x}{1+x^4}dx + \int\frac{1}{1+4y^2}dy = 0$

Hint, for the x variable consider using partial fractions to simplify the equation before you try and integrate.

The second one is a little harder.

$\displaystyle (3x^2 + 9xy + 5y^2)dx - (6x^2 +4xy)dy = 0$

For this one I would suggest using a substitution to simplify the equation a little before you attempt to separate the variables.

Let $\displaystyle y = vx$ where $\displaystyle v = v(x)$, if we differentiate this equation with respect to $\displaystyle x$ we get

$\displaystyle dy = xdv + vdx$.

Substitute in these 2 values and see how you go.

When you've done these try and see what you can manage with the last two.

Hope this helps