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Popular >

y^'+y=2e^x,y(0)=20

Solution

y^{^{'}} + y = 2 e^{x}, y (0) = 20

Solution

y = e^{x} + \frac{19}{e ^{x}}

Solution steps

y^{'} (x) + y = 2 e^{x}

Solve linear ODE:

y = e^{x} + \frac{19}{e ^{x}}

y = e^{x} + \frac{19}{e ^{x}}

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Frequently Asked Questions (FAQ)

What is the solution for y^'+y=2e^x,y(0)=20 ?
The solution for y^'+y=2e^x,y(0)=20 is y=e^x+(19)/(e^x)