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Popular >

y^'=xy+8x+5y+40

Solution

y^{^{'}} = x y + 8 x + 5 y + 40

Solution

y = e^{\frac{x ^{2}}{2} + 5 x + c_{1}} - 8

Solution steps

y^{'} (x) = x y + 8 x + 5 y + 40

Solve separable ODE:

y = e^{\frac{x ^{2}}{2} + 5 x + c_{1}} - 8

y = e^{\frac{x ^{2}}{2} + 5 x + c_{1}} - 8

Popular Examples

y^{^{'}} = 2 x^{3} \cdot y^{2}

y*(dy)/(dx)=((x-e^x))/(e^{y^2)}

y \cdot \frac{d y}{d x} = \frac{( x - e ^{x} )}{e ^{y^{2}}}

- y^{^{'}} - 9 y = 0

(dy)/(dx)=x^2(y-2),y(0)=9

\frac{d y}{d x} = x^{2} (y - 2), y (0) = 9

(dy)/(dx)=x^2(y-2),y(0)=8

\frac{d y}{d x} = x^{2} (y - 2), y (0) = 8

Frequently Asked Questions (FAQ)

What is the solution for y^'=xy+8x+5y+40 ?
The solution for y^'=xy+8x+5y+40 is y=e^{(x^2)/2+5x+c_{1}}-8