What is the solution for sqrt(y)dx+(x-4)dy=0 ?

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

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sqrt(y)dx+(x-4)dy=0

y d x + (x - 4) d y = 0

y = \frac{ln ^{2} ( x - 4 )}{4} - \frac{c _{1} ln ( x - 4 )}{2} + \frac{c _{1}^{2}}{4}

Solution steps

y d x + (x - 4) d y = 0

Solve separable ODE:

y = \frac{ln ^{2} ( x - 4 )}{4} - \frac{c _{1} ln ( x - 4 )}{2} + \frac{c _{1}^{2}}{4}

y = \frac{ln ^{2} ( x - 4 )}{4} - \frac{c _{1} ln ( x - 4 )}{2} + \frac{c _{1}^{2}}{4}

\frac{d}{d x} (e^{x^{- 2}})

\frac{\partial}{\partial x} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}})

\frac{4}{( s ^{2} + 4 )}

x \to \infty lim (csc (x))

3 \frac{d y}{d x} = x^{2}

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