What is the solution for y^'=e^{(2+1)x+(1+2)y} ?

The solution for y^'=e^{(2+1)x+(1+2)y} is y=-(ln(-e^{3x}-c_{1}))/3

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y^'=e^{(2+1)x+(1+2)y}

y^{^{'}} = e^{(2 + 1) x + (1 + 2) y}

y = - \frac{ln ( - e ^{3 x} - c _{1} )}{3}

Solution steps

y^{'} (x) = e^{(2 + 1) x + (1 + 2) y}

Solve separable ODE:

y = - \frac{ln ( - e ^{3 x} - c _{1} )}{3}

y = - \frac{ln ( - e ^{3 x} - c _{1} )}{3}

0.001 \cdot y^{^{'}} - \frac{y}{0.001} = 0

y (x^{2} + 3) d y = 2 x (y^{2} - 1) d x

\frac{d p}{d t} = k (600 - p)

x^{^{'}} = - 100 x, x (0) = 1

\frac{d y}{d x} = 3 3 y^{2}, y (0) = 0

What is the solution for y^'=e^{(2+1)x+(1+2)y} ?
The solution for y^'=e^{(2+1)x+(1+2)y} is y=-(ln(-e^{3x}-c_{1}))/3