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Popular >

y^'=2x+y,y(0)=6

Solution

y^{^{'}} = 2 x + y, y (0) = 6

Solution

y = - 2 (x + 1) + 8 e^{x}

Solution steps

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

Solve linear ODE:

y = - 2 (x + 1) + 8 e^{x}

y = - 2 (x + 1) + 8 e^{x}

Popular Examples

area y=x^2+1,y=x+3 area

y = x^{2} + 1, y = x + 3

(x + 2 y) y^{^{'}} = 5 x - y

integral of (sin(t))/t

\int \frac{sin ( t )}{t} d t

derivative of 4/(1+4x)

\frac{d}{d x} (\frac{4}{1 + 4 x})

integral of x*sec^2(x)

\int x \cdot sec^{2} (x) d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=2x+y,y(0)=6 ?
The solution for y^'=2x+y,y(0)=6 is y=-2(x+1)+8e^x