How do you solve the following differential equation?

dy/dt y(t) = 1

Use the separation of variables method:



ydy = dt





(1/2)y2 = t + c





y2 = 2t + c





y = ±√{2t + c}





You need a boundary condition to determine whether it is + or - and the value of c.

y dy = dt

(1/2)y^2 = t + C1

y = +/- sqrt[2t + C]

The meaning of choosing "C1" as the constant of integration and then changing it to "C"

is that the original constant of integration got multiplied by 2 between lines 2 and 3...

but of course it's still a constant.

dy/dt=y'(t) ; so your question is y'(t)/y(t)=1 so we can make y'(t).dt/y(t)=dt then we integrate the both side; int( y'(t).dt/y(t) ) = int (dt) = t . then say u = y(t) and ln(u) = t , ln (y(t) ) = t ; y = e^t .

y[t] = Sqrt[2] Sqrt[t + C[1]]

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How do you solve the following differential equation?

dy/dt y(t) = 1