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In: Advanced Math

t^2 y'' − 4ty' + 6y = t^4*e^t , t > 0. Use variation of parameters...

t^2 y'' − 4ty' + 6y = t^4*e^t , t > 0. Use variation of parameters to find a particular solution given that y1 = t^2 and y2 = t^3 are a fundamental set of solutions to the corresponding homogeneous equation

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Question : y'''+4y' =0 , y'''-2y''+4y'-8y=0 , y'''-3y''+3y'-y=0 , y^4 -4y'''+6y''-4y+y=0 , y^4+6y''+9y=0 , y^6+y'''=0
Question : y'''+4y' =0 , y'''-2y''+4y'-8y=0 , y'''-3y''+3y'-y=0 , y^4 -4y'''+6y''-4y+y=0 , y^4+6y''+9y=0 , y^6+y'''=0
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