Final Answer
$\frac{1}{2}\arctan\left(2y\right)=\frac{1}{4}\ln\left(-\sqrt{2}x+1\right)+\frac{1}{4}\ln\left(\sqrt{2}x+1\right)+C_0$
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Step-by-step Solution
Problem to solve:
$\left(1+x^4\right)dy+x\left(1+4y^2\right)dx=0$
Specify the solving method
1
Grouping the terms of the differential equation
$\left(1+x^4\right)dy=0-x\left(1+4y^2\right)dx$
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$\left(1+x^4\right)dy=0-x\left(1+4y^2\right)dx$
Learn how to solve problems step by step online. Solve the differential equation (1+x^4)dy+x(1+4y^2)dx=0. Grouping the terms of the differential equation. x+0=x, where x is any expression. Group the terms of the differential equation. Move the terms of the y variable to the left side, and the terms of the x variable to the right side of the equality. Simplify the expression \frac{-x}{1+x^4}dx.
Final Answer
$\frac{1}{2}\arctan\left(2y\right)=\frac{1}{4}\ln\left(-\sqrt{2}x+1\right)+\frac{1}{4}\ln\left(\sqrt{2}x+1\right)+C_0$