What is the solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=16 ?

The solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=16 is y=\sqrt[3]{3969e^{-3x}+126e^{-(3x)/2}+1}

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y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=16

y^{\frac{1}{2}} \frac{d y}{d x} + y^{\frac{3}{2}} = 1, y (0) = 16

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 ( 63 )}{3}

Isolate

Check solutions by applying initial conditions

\int \frac{sin ( 2 θ )}{1 + sin ^{2} ( θ )} d θ

\int 2 + 4 x^{2} d x

\int e^{x} 1 - e^{x} d x

e^{x^{2} y} = x + y

\frac{\partial}{\partial x} (\frac{x}{x ^{2} - y ^{2}})

What is the solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=16 ?
The solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=16 is y=\sqrt[3]{3969e^{-3x}+126e^{-(3x)/2}+1}