Consider the first order ordinary differential equation with P(t) = Find the gen

Question

Consider the first order ordinary differential equation with P(t) = Find the general solution for 0 t 2. Find the constant in the general solution for 0 t 2 so that the initial condition is satisfied. Find a general solution for t > 2. Find the constant in the general solution for t > 2 so that the solution for t > 2 and the solution for 0 t 2 match at t = 2. Use Maple, Matlab, or MS-excel to plot the solution for 0 t 5. Explain, with reference to your expression for y(t) and plot of y(t), whether or not the solution is continuous at t = 2. Explain, with reference to your expression for y(t) and plot of y(t), whether or not the solution is differentiable at t = 2.

Explanation / Answer

a)

for 0 <= t <= 2

here P(t) = 1

dy/dt + y = t

here Q = t

I.F. = e^integral P dt = e^t

so the solution is

y * e^t = integral[e^t * t]dt + C

or

y = (t - 1) + Ce^-t

b)

y = (t - 1) + Ce^-t

put x = 0 , y = 1

we get

1 = -1 + C

or

C = 2

y = (t - 1) + 2e^-t

c)

for t>2

P(t) = 3

I.F. = e^3t

solution is

y*e^3t = integral[e^3t * t]dt + C

or

y = (t-1)/3 + Ce^-3t

here t = 2

y = 1

1 = 1/3 + Ce^-6

or

C = (2/3)e^6

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