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Let $y=y_1(t)$ be a solution of $y'+p(t)y=0$ and let $ y=y_2(t)$ be a solution of $y'+p(t)y=g(t)$. How can we show that $ y=y_1(t)+y_2(t)$ is also a solution of $y'+p(t)y=g(t)$?

I'm not really sure how to approach the problem. It's in the section dealing with differences between linear and nonlinear equations.

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1 Answer 1

$(y_1+y_2)'(t)+p(t)(y_1+y_2)(t)$

$=y_1'(t)+y_2'(t)+p(t)y_1(t)+p(t)y_2(t)$

$=y_1'(t)+p(t)y_1(t)+y_2'(t)+p(t)y_2(t)$

$=0+g(t)$

$=g(t)$

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