What is the solution for y^'+1/(200+t)y=0,y(0)=60 ?

The solution for y^'+1/(200+t)y=0,y(0)=60 is y=(12000)/(t+200)

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y^'+1/(200+t)y=0,y(0)=60

y^{^{'}} + \frac{1}{200 + t} y = 0, y (0) = 60

y = \frac{12000}{t + 200}

Solution steps

y^{'} (t) + \frac{1}{200 + t} y = 0

Solve separable ODE:

y = \frac{12000}{t + 200}

y = \frac{12000}{t + 200}

What is the solution for y^'+1/(200+t)y=0,y(0)=60 ?
The solution for y^'+1/(200+t)y=0,y(0)=60 is y=(12000)/(t+200)