# Thread: Differential equations - seperation of variables

1. ## 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

2. 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

For the first one:

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

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

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

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.

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

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

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

$dy = xdv + vdx$.

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

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# dy/dQ rsecQ=cosQ

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