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Popular >

y^'+xy=x

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

tangent of y=7e^{1-x^2},\at tangent of

y = 7 e^{1 - x^{2}}, at

derivative of (1-x/2 ^{e^{x^2}})

\frac{d}{d x} ((1 - \frac{x}{2})^{e^{x^{2}}})

x(dy)/(dx)+(3x+1)y=e^{3x}

x \frac{d y}{d x} + (3 x + 1) y = e^{3 x}

(6y+2t-7)dt+(8y+6t-1)dy=0

(6 y + 2 t - 7) d t + (8 y + 6 t - 1) d y = 0

integral of x/(sqrt(3-2x-x^2))

\int \frac{x}{3 - 2 x - x ^{2}} d x

Frequently Asked Questions (FAQ)

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