Question #236266

8)Solve (D²+3DD'+2D'^2)z=x cos y+e^x+y, Where D=∂\∂x and D'=∂\∂y.

Explain the problem with step by step process?


Expert's answer

The degree of the differential equation is the power of the highest order derivative, where the original equation is represented in the form of a polynomial equation in derivatives such as y’,y”, y”’, and so on.

(d2y/dx2)+2(dy/dx)+y=0(d2y/dx2)+ 2 (dy/dx)+y = 0 , so the degree of this equation here is 1. See some more examples here:

  • dy/dx+1=0,dy/dx + 1 = 0, degree is 1
  • (y”’)3+3y+6y’–12=0,(y”’)^3 + 3y” + 6y’ – 12 = 0, degree is 3
  • (dy/dx)+cos(dy/dx)=0;(dy/dx) + cos(dy/dx) = 0; it is not a polynomial equation in y′ and the degree of such a differential equation can not be defined.

Note:

Order and degree (if defined) of a differential equation are always positive integers.


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