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Popular >

y^'=2x+y,y(0)=7

Solution

y^{^{'}} = 2 x + y, y (0) = 7

Solution

y = - 2 (x + 1) + 9 e^{x}

Solution steps

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

Solve linear ODE:

y = - 2 (x + 1) + 9 e^{x}

y = - 2 (x + 1) + 9 e^{x}

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x^2y^{''}-xy^'+y= 1/x

x^{2} y^{^{''}} - x y^{^{'}} + y = \frac{1}{x}

integral from 0 to 0.5 of 500x

\int_{0}^{0.5} 500 x d x

tangent of f(x)=6x^2+2x-7 tangent of

f (x) = 6 x^{2} + 2 x - 7

Frequently Asked Questions (FAQ)

What is the solution for y^'=2x+y,y(0)=7 ?
The solution for y^'=2x+y,y(0)=7 is y=-2(x+1)+9e^x