What is the solution for yy^'-4y^'-5y=125 ?

The solution for yy^'-4y^'-5y=125 is y=-29\W(-1/(29e^{(5t+c_{1))/(29)}})-25

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yy^'-4y^'-5y=125

y y^{^{'}} - 4 y^{^{'}} - 5 y = 125

y = - 29 W (- \frac{1}{29 e ^{\frac{5 t + c _{1}}{29}}}) - 25

Solution steps

y y^{'} (t) - 4 y^{'} (t) - 5 y = 125

Solve separable ODE:

y = - 29 W (- \frac{1}{29 e ^{\frac{5 t + c _{1}}{29}}}) - 25

y = - 29 W (- \frac{1}{29 e ^{\frac{5 t + c _{1}}{29}}}) - 25

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

3 (16 + y^{2}) d x + 2 x y d y = 0

u^{^{'}} - r H (t) + r u (t)^{2} \frac{1}{k} = 0

(y + x y) + (x - x y) y^{^{'}} = 0

\frac{d y}{d x} + y = y x e^{x + 8}

What is the solution for yy^'-4y^'-5y=125 ?
The solution for yy^'-4y^'-5y=125 is y=-29\W(-1/(29e^{(5t+c_{1))/(29)}})-25