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Let y(x) be the solution of the initial value problem: y′+2y=xe-2x, y(1)=0. What is y(−1), correct...

Let y(x) be the solution of the initial value problem:

y′+2y=xe-2x, y(1)=0.

What is y(−1), correct to 1 decimal place?

Math   Calculus   
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