What is the solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=4 ?

The solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=4 is y=\sqrt[3]{49e^{-3x}+14e^{-(3x)/2}+1}

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y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=4

y^{\frac{1}{2}} \frac{d y}{d x} + y^{\frac{3}{2}} = 1, y (0) = 4

y = 3 49 e^{- 3 x} + 14 e^{- \frac{3 x}{2}} + 1

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 ( 7 )}{3}

Isolate

y : y = 3 49 e^{- 3 x} + 14 e^{- \frac{3 x}{2}} + 1

Check solutions by applying initial conditions

y = 3 49 e^{- 3 x} + 14 e^{- \frac{3 x}{2}} + 1

y = x^{2} - 8 x

\int \frac{5 + 7 x}{1 + x ^{2}} d x

\frac{d}{d x} (4 + 3 x^{2})

\int (x - 8)^{2} d x

\frac{3}{s} (1 - e^{- s})

What is the solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=4 ?
The solution for y^{1/2}(dy)/(dx)+y^{3/2}=1,y(0)=4 is y=\sqrt[3]{49e^{-3x}+14e^{-(3x)/2}+1}