What is the solution for y^{1/2}((dy)/(dx))+y^{3/2}=1,y(0)=9 ?

The solution for y^{1/2}((dy)/(dx))+y^{3/2}=1,y(0)=9 is y=\sqrt[3]{676e^{-3x}+52e^{-(3x)/2}+1}

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y^{1/2}((dy)/(dx))+y^{3/2}=1,y(0)=9

y^{\frac{1}{2}} (\frac{d y}{d x}) + y^{\frac{3}{2}} = 1, y (0) = 9

Solution steps

y^{\frac{1}{2}} (\frac{d y}{d x}) + y^{\frac{3}{2}} = 1

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{\frac{1}{2}} (y^{^{'}} (x)) + y^{\frac{3}{2}} = 1

Solve for

x, y^{'} (x) = \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{x ^{^{'}}}

y^{\frac{1}{2}} ((\frac{1}{x ^{^{^{'}}}})) + y^{\frac{3}{2}} = 1

Isolate

x^{^{'}} : x^{^{'}} = - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

x = \int - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1} d y

\int - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1} d y = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + c_{1}

x = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + c_{1}

Apply initial conditions:

x = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + \frac{2 ln ( 26 )}{3}

Isolate

Check solutions by applying initial conditions

x \to 0 lim (e^{- \frac{2}{x ^{3}}})

f (x) = (x^{3} + 7) e^{x}

\int \frac{2 x}{x ^{2} + 14 x + 49} d x

n = 1 \sum \infty 5 \cdot \frac{2}{n}

x \to 6 lim (\frac{x ^{2} - 9 x + 18}{x - 6})

What is the solution for y^{1/2}((dy)/(dx))+y^{3/2}=1,y(0)=9 ?
The solution for y^{1/2}((dy)/(dx))+y^{3/2}=1,y(0)=9 is y=\sqrt[3]{676e^{-3x}+52e^{-(3x)/2}+1}