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