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Popular >

y^'=x+y

Solution

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

Solution

y = - x - 1 + c_{1} e^{x}

Solution steps

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

Solve linear ODE:

y = - x - 1 + c_{1} e^{x}

y = - x - 1 + c_{1} e^{x}

Graph

Popular Examples

y^{''}+y=sin(6t),y(0)=0,y^'(0)=0

y^{^{''}} + y = sin (6 t), y (0) = 0, y^{^{'}} (0) = 0

(\partial)/(\partial x)(1+cos^4(x^4))

\frac{\partial}{\partial x} (1 + cos^{4} (x^{4}))

derivative of sin(ln(w))derivative of

sin (ln (w))

(4y+2t-5)dt+(6y+4t-1)dy=0

(4 y + 2 t - 5) d t + (6 y + 4 t - 1) d y = 0

sum from n=0 to infinity of e^{-an}

n = 0 \sum \infty e^{- an}

Frequently Asked Questions (FAQ)

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