Antiderivatives

**sw3etazngyrl** · June 5th 2007, 08:34 PM

How do you find the antiderivatives of:

1. x+4y^(3)*[(x^(2)+1)^2]*dy/dx=0

2. (y^2)/(x^6)=dy/dx

3. y^2+1=dy/dx

4. y'=(sinx)/(siny)

Thanks!

**curvature** · June 5th 2007, 10:06 PM

Well, they are some differential equations.
In fact,they are all separable differential equations.

**earboth** · June 6th 2007, 01:25 AM

Originally Posted by sw3etazngyrl

How do you find the antiderivatives of:

1. x+4y^(3)*[(x^(2)+1)^2]*dy/dx=0

...

Hello,

$x+4y^3 \cdot (x^2+1)^2 \cdot \frac{dy}{dx}=0\; \Longleftrightarrow \; 4y^3 \cdot dy = -\frac{x}{(x^2+1)^2} \cdot dx$

$\int 4y^3 \cdot dy = \int \left(-\frac{x}{(x^2+1)^2} \right) \cdot dx$ For the RHS use integration by substitution:

$y^4 = \frac{1}{2(x^2+1)} + C \; \Longrightarrow \; y = \left|\sqrt[4]{\frac{1}{2(x^2+1)} + C}\; \; \right|$

**^_^Engineer_Adam^_^** · June 6th 2007, 04:45 AM

Originally Posted by earboth

Hello,

$x+4y^3 \cdot (x^2+1)^2 \cdot \frac{dy}{dx}=0\; \Longleftrightarrow \; 4y^3 \cdot dy = -\frac{x}{(x^2+1)^2} \cdot dx$

$\int 4y^3 \cdot dy = \int \left(-\frac{x}{(x^2+1)^2} \right) \cdot dx$ For the RHS use integration by substitution:

$y^4 = \frac{1}{2(x^2+1)} + C \; \Longrightarrow \; y = \left|\sqrt[4]{\frac{1}{2(x^2+1)} + C}\; \; \right|$

hi
was wondering if there are any constant?
edit
ok its fixed

**curvature** · June 6th 2007, 04:57 AM

the solution of the 1st question again

Thread: Antiderivatives

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