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
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