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Answer to Question #66659 in Differential Equations for devangi
Question #66659
true/false.justify.
The normal form of the differential equation
y" - 4xy' + (4x^2 - 1)y =-3e^x^2 sin2x is
(d^2v/dx^2)+v = -3 sin2x, where v= y e^-x^2
The normal form of the differential equation
y" - 4xy' + (4x^2 - 1)y =-3e^x^2 sin2x is
(d^2v/dx^2)+v = -3 sin2x, where v= y e^-x^2
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Thank you for correcting us. The file was updated.
wrong solution plzz upload correct solution
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