Linear differential equations sometimes occur in which one or both of the functi

Q: Linear differential equations sometimes occur in which one or both of the functi

here t0 =1 for 0 d(e^6x .y)/dx = e^6x => e^6x.y = e^6x/6 +c => y = 1/6 + c e-6x given y(0) = 0 => c = -1/6 and hence y1 = 1/6 ( 1- e-6x) for t>t0 when g(t) = 0 y' +6y = 0 => y2 = ce-6x since we have to make it continous at t= t0 =1. then y1(1) = y2(1) => c = 1/6(e-6 -1) hence y2 = 1/6(e6 -1)e-6x So final solution is y = 1/6 ( 1- e-6x) when 0

ID: 1948610 • Letter: L

Question

Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(t) y = g(t) have jump discontinuities. If t0 is such a point of discontinuity, then it is necessary to solve the equation separately for tt0. Afterward, the two solutions are matched so that y is continuous at t0 ; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at t0. Solve the initial value problem. y' + 6y = g(t), y(0) = 0

Explanation / Answer

here t0 =1

for 0<t<t0 when g(t) =1

y' + 6y = 1

=> e^6x.y' + 6e^6x.y = e^6x [ on multiplying by e^(6x)]

=> d(e^6x .y)/dx = e^6x => e^6x.y = e^6x/6 +c => y = 1/6 + c e-6x

given y(0) = 0 => c = -1/6 and hence

y1 = 1/6 ( 1- e-6x)

for t>t0 when g(t) = 0

y' +6y = 0

=> y2 = ce-6x

since we have to make it continous at t= t0 =1.

then y1(1) = y2(1) => c = 1/6(e-6 -1)

hence y2 = 1/6(e6 -1)e-6x

So final solution is

y = 1/6 ( 1- e-6x) when 0<t<1

= 1/6(e6 -1)e-6x when t>1

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Linear differential equations sometimes occur in which one or both of the functi

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