The question is that find the general solution of differential equation y″−2y′+y=exi know that yc=Axex+Bexthen let the f(x)=ex, so y=pxex as f(x) is in the complementary function. So y′=pex(x+1)y″=pex(x+2) then i put it into equation,but the answer is (0)pex=ex,then i can't find the unknown number p.

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The question is that find the general solution of differential equation y″−2y′+y=exi know that yc=Axex+Bexthen let the f(x)=ex, so y=pxex as f(x) is in the complementary function. So y′=pex(x+1)y″=pex(x+2) then i put it into equation,but the answer is (0)pex=ex,then i can&#039;t find the unknown number p.

einfachmoipf · Accepted Answer

Youre

Mason Hall · Accepted Answer

As a quick method of solving this problem quickly, note that multiplying both sides by e−x yieldse−xy″−2e−xy′+e−xy=(e−xy)″=1To see this you could note thaty″−2y′+y=(y′−y)′−(y′−y)for which the appropriate integrating factor would be e−x. And after integrating, y′−y has the same integrating factor. Or it&#039;s easy to show that(uv)″=u″v+2u′v′+uv″Once in the appropriate form, all that is needed is to integrate both sides twice to yielde−xy=12x2+ax+b, y=(12x2+ax+b)ex

The question is that find the general solution of differential

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