Question
Asked Nov 8, 2019
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solve using the Initial Value Problem. 

xy' + y = ex ,       y(1) = 3

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Expert Answer

Step 1

Given,

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e and y(1) = 3 xy'

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Step 2

Now,

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ху' + у %3D е* ex у — у' + х х The above expression is in the form of y'(x) p(x)y = q(x) So, on comparing we get p(x) ex and q(x) х Now Integrating factor, I(x) = es P(x)dx = e#dx enx = X

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Step 3
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y TeIdx . + c] [/ 1(х)q(x)dx + c] 1(х) (dx х х =Se*dx + c] ex+c х Since, y(1) 3 e+c 3 1 ec 3 — с %3D 3 — е eX+3-e :. у- х

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Tagged in

Math

Calculus

Differential Equations