Question

In Exercises 31-42, solve the initial value problem.

3(x^(2))y''-4xy'+2y=0, y(1)=2, y'(1)=1

Answer #1

Solve the Initial Value Problem:
?x′ = 2y−x
y′ = 5x−y
Initial Conditions:
x(0)=2
y(0)=1

For the initial value problem
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The solution is y=____.

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x′=x+y
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x(0) = 0, y(0) = 0.

(x2+2)y^'' + 4xy^' + 2y = 0 , y(0) =0 , y^'(0) =
1
Solve by using power series around x0= 0

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y = 0; y(0)=0, y'(0)=1

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y'(0)=0

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