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Superposition Principle. Lety1be a solution to y''(t)+p(t)y'(t)+q(t)y(t)=g1(t) on the interval Iand lety2be a solution to y''(t)+p(t)y'(t)+q(t)y(t)=g2(t) on the same interval. Show that for any constantsk1and k2, the functionk1y1+k2y2is a solution on Ito y''(t)+p(t)y'(t)+q(t)y(t)=k1g1(t)+k2g2(t).

Short Answer

Expert verified

k1y1+k2y2is the solution to the y''(t)+p(t)y'(t)+q(t)y(t)=k1g1(t)+k2g2(t).

Step by step solution

01

Check whether or be the solution

If y1be the solution of the differential equationy''(t)+p(t)y'(t)+q(t)y(t)=g1(t)

Then y1''(t)+p(t)y1'(t)+q(t)y1(t)=g1(t)and if y2be the solution of the differential equation;

y''(t)+p(t)y'(t)+q(t)y(t)=g2(t)

Theny2''(t)+p(t)y2'(t)+q(t)y2(t)=g2(t)

So, let's take y=ky1+ky2theny'=k1y1'+k2y2'y''=k1y1''+k2y2''

02

Substitute the values

Substitute these equations in the differential equation:

y''(t)+p(t)y'(t)+q(t)y(t)=k1y1''+k2y2''+p(t)k1y1'+k2y2'+q(t)ky1+ky2=k1y1''(t)+p(t)y1'(t)+q(t)y1(t)+k2y2''(t)+p(t)y2'(t)+q(t)y2(t)

Then from (1) and (2)

y''(t)+p(t)y'(t)+q(t)y(t)=k1g1(t)+k2g2(t)

Therefore k1y1+k2y2is the solutions to they''(t)+p(t)y'(t)+q(t)y(t)=k1g1(t)+k2g2(t)

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