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Popular >

y^'=x+y,y(0)=1

Solution

y^{^{'}} = x + y, y (0) = 1

Solution

y = - x - 1 + 2 e^{x}

Solution steps

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

Solve linear ODE:

y = - x - 1 + 2 e^{x}

y = - x - 1 + 2 e^{x}

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\int_{0}^{R} \frac{1}{1 + x ^{2}} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=x+y,y(0)=1 ?
The solution for y^'=x+y,y(0)=1 is y=-x-1+2e^x