Find a particular solution of y"+y=2e^x

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Find a particular solution of y"+y=2e^x

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Use the annihilator method on the 2e^x. Think, what differential operator would result in the answer.
is there a formula i can follow? this is supposed to be a non homogeneous second order differential equation.
Well, you need to use the differential operator concept. What D would destroy 2e^x? Then use that differential operator and create the particular solution. Then plug that solution into the function, and see what the coefficient is.

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(e^x)"+(e^x)=2e^x
D-1 = Ae^(1x) yp = Ae^(1x) e^x = Ae^x+A^ex = 2e^x
sorry, function, not e^x
also, use t for 1.
I mangled that a little bit. Your y_p = Ae^x. You have to determine what A is. To do that, substitute the y_p back into the function, and solve for A.

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