What is the solution for y^'=-0.02y+3,y(0)=0 ?

The solution for y^'=-0.02y+3,y(0)=0 is y=150-(150)/(e^{0.02t)}

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y^'=-0.02y+3,y(0)=0

y^{^{'}} = - 0.02 y + 3, y (0) = 0

y = 150 - \frac{150}{e ^{0.02 t}}

Solution steps

y^{'} (t) = - 0.02 y + 3

Solve linear ODE:

y = 150 - \frac{150}{e ^{0.02 t}}

y = 150 - \frac{150}{e ^{0.02 t}}

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

\frac{d y}{d x} = y x, y (1) = 3

2 sin (x) - y^{2} \frac{d y}{d x} = 0

2 x y^{3} d x - (1 + x^{2}) d y = 0

\frac{d y}{d x} = 0.04 - \frac{y}{7}

What is the solution for y^'=-0.02y+3,y(0)=0 ?
The solution for y^'=-0.02y+3,y(0)=0 is y=150-(150)/(e^{0.02t)}