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

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

Check solutions by applying initial conditions

\frac{d}{d x} (ln (x + y^{2}))

\frac{d}{d x} (ln ∣ sec (x) + tan (x) ∣)

x \frac{d y}{d x} = y^{2} - y, y (\frac{1}{4}) = \frac{1}{4}

x^{2} + a x + b

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{7}})

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}