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

The solution for y^{1/2}y^'+y^{3/2}=1,y(0)=4 is y=\sqrt[3]{49e^{-3t}+14e^{-(3t)/2}+1}

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

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

Solution steps

y^{\frac{1}{2}} y^{'} (t) + y^{\frac{3}{2}} = 1

Solve for

t, y^{'} (t) = \frac{d y}{d t} = \frac{1}{\frac{d t}{d y}} = \frac{1}{t ^{^{'}}}

y^{\frac{1}{2}} (\frac{1}{t ^{^{^{'}}}}) + y^{\frac{3}{2}} = 1

Isolate

t^{^{'}} : t^{^{'}} = - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1}

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

then

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

t = \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}

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

Apply initial conditions:

t = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + \frac{2 ln ( 7 )}{3}

Isolate

Check solutions by applying initial conditions

\frac{\partial}{\partial x} (x \cdot y \cdot z)

y = x^{4} + 7 e^{x}, (0, 7)

\int \frac{x}{9 + x ^{4}} d x

ln (16 sin^{2} (x))

x \to 7 + lim (\frac{1}{( x - 7 ) ^{2}})

What is the solution for y^{1/2}y^'+y^{3/2}=1,y(0)=4 ?
The solution for y^{1/2}y^'+y^{3/2}=1,y(0)=4 is y=\sqrt[3]{49e^{-3t}+14e^{-(3t)/2}+1}