What is the solution for y^'=x^4*y^{-1},y(0)=7 ?

The solution for y^'=x^4*y^{-1},y(0)=7 is y=sqrt((2x^5)/5+49)

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y^'=x^4*y^{-1},y(0)=7

y^{^{'}} = x^{4} \cdot y^{- 1}, y (0) = 7

y = \frac{2 x ^{5}}{5} + 49

Solution steps

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

Solve separable ODE:

y = \frac{2 x ^{5}}{5} + 49

y = \frac{2 x ^{5}}{5} + 49

(cos (θ)) d x + (1 + e - x) (sin (θ)) d θ = 0

\frac{d r}{d q} = (50 - 4 q) e^{- \frac{r}{5}}

\frac{d r ( θ )}{r ( θ ) ^{2}} = \frac{r ( θ ) d θ}{θ ^{2}}

x y d x + x^{2} d y = 0

- 9 y^{^{'}} + 3 y = 8

What is the solution for y^'=x^4*y^{-1},y(0)=7 ?
The solution for y^'=x^4*y^{-1},y(0)=7 is y=sqrt((2x^5)/5+49)