What is the solution for x^2+4xy+x((dy)/(dx))=0 ?

The solution for x^2+4xy+x((dy)/(dx))=0 is y=-(4x-1)/(16)+(c_{1})/(e^{4x)}

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x^2+4xy+x((dy)/(dx))=0

x^{2} + 4 x y + x (\frac{d y}{d x}) = 0

y = - \frac{4 x - 1}{16} + \frac{c _{1}}{e ^{4 x}}

Solution steps

x^{2} + 4 x y + x (\frac{d y}{d x}) = 0

Solve linear ODE:

y = - \frac{4 x - 1}{16} + \frac{c _{1}}{e ^{4 x}}

y = - \frac{4 x - 1}{16} + \frac{c _{1}}{e ^{4 x}}

\frac{\partial}{\partial x} (y ln (5 x + y))

\int_{1}^{6} 7 \frac{ln ( x )}{x ^{2}} d x

\int sin (2 x) e^{c o s^{2} (x)} d x

x \to 3 lim (\frac{6}{x - 3})

2 x - 3

What is the solution for x^2+4xy+x((dy)/(dx))=0 ?
The solution for x^2+4xy+x((dy)/(dx))=0 is y=-(4x-1)/(16)+(c_{1})/(e^{4x)}