What is the solution for f(x)=f^'(x)(4x+9)^5 ?

The solution for f(x)=f^'(x)(4x+9)^5 is f(x)=e^{-1/(16(4x+9)^4)+c_{1}}

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f(x)=f^'(x)(4x+9)^5

f (x) = f^{^{'}} (x) (4 x + 9)^{5}

f (x) = e^{- \frac{1}{16 ( 4 x + 9 ) ^{4}} + c_{1}}

Solution steps

f (x) = f^{'} (x) (4 x + 9)^{5}

Solve separable ODE:

f (x) = e^{- \frac{1}{16 ( 4 x + 9 ) ^{4}} + c_{1}}

f (x) = e^{- \frac{1}{16 ( 4 x + 9 ) ^{4}} + c_{1}}

\frac{d y}{d x} = - 7 (- 4 - y)

\frac{d h}{d t} = - \frac{1}{72} h

(y) sin (x) + y^{^{'}} = sin (x)

x^{2} y - 3 x y^{^{'}} - 2 y = 0

sin (x) d x + \frac{1}{y} d y = 0, y (0) = 0

What is the solution for f(x)=f^'(x)(4x+9)^5 ?
The solution for f(x)=f^'(x)(4x+9)^5 is f(x)=e^{-1/(16(4x+9)^4)+c_{1}}