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Popular >

y^'+ty=y

Solution

y^{^{'}} + t y = y

Solution

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

Solution steps

y^{'} (t) + t y = y

Solve separable ODE:

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

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

Popular Examples

x^2(dy)/(dx)+xy-y=0

x^{2} \frac{d y}{d x} + x y - y = 0

(1+x^2)y^'-xy=0

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

2xtan(y)dx+sec^2(y)dy=0

2 x tan (y) d x + sec^{2} (y) d y = 0

u^'=csc(\H(x))cos(x/2)

u^{^{'}} = csc (H (x)) cos (\frac{x}{2})

y^'=e^xcos^2(y),y(0)= pi/4

y^{^{'}} = e^{x} cos^{2} (y), y (0) = \frac{π}{4}

Frequently Asked Questions (FAQ)

What is the solution for y^'+ty=y ?
The solution for y^'+ty=y is y=e^{t-(t^2)/2+c_{1}}