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

y = 3 676 e^{- 3 x} + 52 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 ( 26 )}{3}

Isolate

y : y = 3 676 e^{- 3 x} + 52 e^{- \frac{3 x}{2}} + 1

Check solutions by applying initial conditions

y = 3 676 e^{- 3 x} + 52 e^{- \frac{3 x}{2}} + 1

f (x) = cos (x) - 1

\frac{d}{d x} (e^{2 x})

\int \frac{y}{( y + 2 ) ( 5 y - 1 )} d y

y^{^{''}} + y^{^{'}} = sin (t)

\int \frac{1}{( x \cdot x ^{2} - 16 )} d x

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}