$\left(1+x^4\right)\cdot dy+x\cdot\left(1+4y^2\right)\cdot dx=0$
$\frac{dy}{dx}=\frac{2x}{3y^2}$
$\frac{dx}{dy}=y+xy$
$dx+e^{3x}dy=0$
$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$
$\frac{dy}{dx}=e^{3x+2y}$
$y'=x+8$
Differential Equations
~ 0.06 s
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