What is the solution for y^'=13xy-16x ?

The solution for y^'=13xy-16x is y=(e^{(13x^2+c_{1})/2})/(13)+16/13

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y^'=13xy-16x

y^{^{'}} = 13 x y - 16 x

y = \frac{e ^{\frac{13 x ^{2} + c _{1}}{2}}}{13} + \frac{16}{13}

Solution steps

y^{'} (x) = 13 x y - 16 x

Solve separable ODE:

y = \frac{e ^{\frac{13 x ^{2} + c _{1}}{2}}}{13} + \frac{16}{13}

y = \frac{e ^{\frac{13 x ^{2} + c _{1}}{2}}}{13} + \frac{16}{13}

\int \frac{( 2 x ^{3} + x - 1 )}{x ^{3} + x ^{2} - 4 x - 4} d x

x \to 2 lim (π^{2} - 2 x + 5)

x (2 x + 7)^{\frac{2}{3}}

t \to 2 lim (8 + t^{3})

y = e^{- x} sin (x), y = - e^{- x} sin (x), 0 \leq x \leq π

What is the solution for y^'=13xy-16x ?
The solution for y^'=13xy-16x is y=(e^{(13x^2+c_{1})/2})/(13)+16/13