What is the solution for ln((dy)/(dx))=ax+by ?

The solution for ln((dy)/(dx))=ax+by is y=-(ln(-(be^{ax})/a-c_{1}b))/(b)

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ln((dy)/(dx))=ax+by

ln (\frac{d y}{d x}) = a x + b y

y = - \frac{ln ( - \frac{b e ^{a x}}{a} - c _{1} b )}{b}

Solution steps

ln (\frac{d y}{d x}) = a x + b y (x)

Solve separable ODE:

y = - \frac{ln ( - \frac{b e ^{a x}}{a} - c _{1} b )}{b}

y = - \frac{ln ( - \frac{b e ^{a x}}{a} - c _{1} b )}{b}

2 y y^{^{'}} = 15 t^{2}, y (0) = 4

\frac{d y}{d x} = - \frac{7}{4} x^{3} y^{3}

x^{2} d y = (9 + y^{2}) d x

(1 - y^{2})^{\frac{1}{2}} d x - (1 - x^{2})^{\frac{1}{2}} d y = 0

\frac{d y}{d x} = y sin (3 x) - 3 y e^{- 3 x}, y (0) = 1

What is the solution for ln((dy)/(dx))=ax+by ?
The solution for ln((dy)/(dx))=ax+by is y=-(ln(-(be^{ax})/a-c_{1}b))/(b)