What is the solution for y^'=1-(2y)/(100+2x),y(0)=20 ?

The solution for y^'=1-(2y)/(100+2x),y(0)=20 is y=(50x)/(50+x)+(x^2)/(2(50+x))+(1000)/(50+x)

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y^'=1-(2y)/(100+2x),y(0)=20

y^{^{'}} = 1 - \frac{2 y}{100 + 2 x}, y (0) = 20

y = \frac{50 x}{50 + x} + \frac{x ^{2}}{2 ( 50 + x )} + \frac{1000}{50 + x}

Solution steps

y^{'} (x) = 1 - \frac{2 y}{100 + 2 x}

Solve linear ODE:

y = \frac{50 x}{50 + x} + \frac{x ^{2}}{2 ( 50 + x )} + \frac{1000}{50 + x}

y = \frac{50 x}{50 + x} + \frac{x ^{2}}{2 ( 50 + x )} + \frac{1000}{50 + x}

t^{2} y^{^{'}} + y = 0

x \to 4 lim (x)

\frac{s + 3}{( s + 3 ) ^{2} + 4}

\frac{d}{d x} (\frac{1}{2} cos (\frac{x}{2}))

y^{^{'}} = y^{2} + y

What is the solution for y^'=1-(2y)/(100+2x),y(0)=20 ?
The solution for y^'=1-(2y)/(100+2x),y(0)=20 is y=(50x)/(50+x)+(x^2)/(2(50+x))+(1000)/(50+x)