What is the solution for y^'=9xsqrt(7y+8),y(0)=3 ?

The solution for y^'=9xsqrt(7y+8),y(0)=3 is y=(567x^4}{16}+\frac{9sqrt(29)x^2)/2+3

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y^'=9xsqrt(7y+8),y(0)=3

y^{^{'}} = 9 x 7 y + 8, y (0) = 3

y = \frac{567 x ^{4}}{16} + \frac{9 29 x ^{2}}{2} + 3

Solution steps

y^{'} (x) = 9 x 7 y + 8

Solve separable ODE:

y = \frac{567 x ^{4}}{16} + \frac{9 29 x ^{2}}{2} + 3

y = \frac{567 x ^{4}}{16} + \frac{9 29 x ^{2}}{2} + 3

\frac{d}{d x} (\frac{x \cdot cos ( x ) + π}{π - x})

\frac{t}{3}

\int (sin^{4} (3 x) - cos^{4} (3 x)) d x

\int r^{3} 1 + 4 r^{2} d r

\frac{1}{3 ( 1 + e ^{- t} ) ^{3}}

What is the solution for y^'=9xsqrt(7y+8),y(0)=3 ?
The solution for y^'=9xsqrt(7y+8),y(0)=3 is y=(567x^4}{16}+\frac{9sqrt(29)x^2)/2+3