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Popular >

y^'=2x+y

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

(3y^3+2xy^3)+(3xy^2+3y^2)((dy)/(dx))=0

(3 y^{3} + 2 x y^{3}) + (3 x y^{2} + 3 y^{2}) (\frac{d y}{d x}) = 0

derivative of (10)/(ln(x))derivative of

\frac{10}{ln ( x )}

area 16-x^2,x^2-2,-3,3 area

16 - x^{2}, x^{2} - 2, - 3, 3

derivative of f(x)=log_{9}(x)derivative of

f (x) = lo g_{9} (x)

derivative of 4/3 x^3-4x^2-60x+10

\frac{d}{d x} (\frac{4}{3} x^{3} - 4 x^{2} - 60 x + 10)

Frequently Asked Questions (FAQ)

What is the solution for y^'=2x+y ?
The solution for y^'=2x+y is y=-2(x+1)+c_{1}e^x