What is the solution for y^'-3/4 y=x^4(y^1)/3 ?

The solution for y^'-3/4 y=x^4(y^1)/3 is y=e^{1/12 ((4x^5)/5+9x)+c_{1}}

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y^'-3/4 y=x^4(y^1)/3

y^{^{'}} - \frac{3}{4} y = x^{4} \frac{y ^{1}}{3}

y = e^{\frac{1}{12} (\frac{4 x ^{5}}{5} + 9 x) + c_{1}}

Solution steps

y^{'} (x) - \frac{3}{4} y = x^{4} \frac{y ^{1}}{3}

Solve separable ODE:

y = e^{\frac{1}{12} (\frac{4 x ^{5}}{5} + 9 x) + c_{1}}

y = e^{\frac{1}{12} (\frac{4 x ^{5}}{5} + 9 x) + c_{1}}

y^{^{'}} = \frac{( ln ^{4} ( x ) )}{( x \cdot y ) ^{2}}, y (1) = 2

y^{^{'}} + 16 x y = 0

\frac{d y}{d x} + y cos (x) = 8 cos (x), y (0) = 10

(x + 1) d t - (x t - x) d x = 0

\frac{d y}{d x} = e^{- 5 y} cos (4 x)

What is the solution for y^'-3/4 y=x^4(y^1)/3 ?
The solution for y^'-3/4 y=x^4(y^1)/3 is y=e^{1/12 ((4x^5)/5+9x)+c_{1}}