1. ## Antiderivatives

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!

2. ## differential equations

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

3. Originally Posted by sw3etazngyrl
How do you find the antiderivatives of:

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

...
Hello,

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

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

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

4. Originally Posted by earboth
Hello,

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

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

$\displaystyle y^4 = \frac{1}{2(x^2+1)} + C \; \Longrightarrow \; y = \left|\sqrt[4]{\frac{1}{2(x^2+1)} + C}\; \; \right|$
hi
was wondering if there are any constant?
edit
ok its fixed

5. ## No 1

the solution of the 1st question again