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Popular >

y^'-5x=xy

Solution

y^{^{'}} - 5 x = x y

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

y^'+ycos(x)=3cos(x),y(0)=5

y^{^{'}} + y cos (x) = 3 cos (x), y (0) = 5

x(dy)/(dx)-7y=0

x \frac{d y}{d x} - 7 y = 0

e^yy^'=((ln(t))^2)/t ,y(1)=ln(19)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (19)

((dy)/(dx))/y = 3/(\sqrt[5]{2x^7+5x+1)}

y^'=((y+1))/((x-1))

y^{^{'}} = \frac{( y + 1 )}{( x - 1 )}

Frequently Asked Questions (FAQ)

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