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Popular >

y^'=-1-4x+2y,y(0)=5

Solution

y^{^{'}} = - 1 - 4 x + 2 y, y (0) = 5

Solution

y = 2 x + \frac{3}{2} + \frac{7}{2} e^{2 x}

Solution steps

y^{'} (x) = - 1 - 4 x + 2 y

Solve linear ODE:

y = 2 x + \frac{3}{2} + \frac{7}{2} e^{2 x}

y = 2 x + \frac{3}{2} + \frac{7}{2} e^{2 x}

Popular Examples

derivative of 2/((2-x^3))

\frac{d}{d x} (\frac{2}{( 2 - x ) ^{3}})

integral of zcos(xz)

\int z cos (x z) d x

(x - 4)^{^{'}}

derivative of f(x)= 3/(x-2)derivative of

f (x) = \frac{3}{x - 2}

(\partial)/(\partial x)(x^3+y^4)

\frac{\partial}{\partial x} (x^{3} + y^{4})

Frequently Asked Questions (FAQ)

What is the solution for y^'=-1-4x+2y,y(0)=5 ?
The solution for y^'=-1-4x+2y,y(0)=5 is y=2x+3/2+7/2 e^{2x}