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Popular >

y^'=2x+y,y(0)=1

Solution

y^{^{'}} = 2 x + y, y (0) = 1

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Graph

Popular Examples

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n = 1 \sum \infty \frac{n ^{7}}{7 ^{n}}

y^'+5y=8,y(0)=0

y^{^{'}} + 5 y = 8, y (0) = 0

derivative of xsqrt(100-x^2)

\frac{d}{d x} (x 100 - x^{2})

Frequently Asked Questions (FAQ)

What is the solution for y^'=2x+y,y(0)=1 ?
The solution for y^'=2x+y,y(0)=1 is y=-2(x+1)+3e^x