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- Use variation of parameters to find the general solution of y′′−2y′+ 2y=(e^x)csc(x)(x+y-4) dx-(3x-y-4)dy=0; when x=4, y=1 Solve using Coefficient linear in the two variablesIf f(x, y) = −x2 + x3 - y + 4xy + y3, then in the process of obtaining its critical points the equation is obtained: The answers are in the picture: