Answer to Question #147857 in Differential Equations for Harsh mishra

Question

2D^2-D'^2+D=x^2-y

Answer 1

$\text{Given } 2D^2-D^2+D=x^2-y \text{ equivalently we have } D^2+D=x^2-y \\ \text{Then we have:}\\y''+y'+y=x^2 \text { then the auxilliary equation of the D.E will be}\\ m^2+m+1=0 \text{ using quadratic formula }\\ m = \frac{-1±\sqrt{1^2-4(1)(1)}}{2(1)}= \frac{-1±\sqrt{-3}}{2}=\frac{-1±3i}{2}=\frac{-1}{2}±\frac{3i}{2}\\ \text{Then the complementary function of the D.E will be of the form } \\ Y_c(x)=e^{-\frac{x}{2}}(A\cos(\frac{3x}{2}) +B\sin (\frac{3x}{2})) \text{ where A and B are constants }\\\text{The particular integral of the D.E is of the form:}\\ Y_p(x)=Cx^2+Dx+E \\ Y_p'= 2Cx+D ; Y_p''=2C\\ Y_p''+Y_p'+Y_p=x^2\\ \implies 2C+2Cx+D +Cx^2+Dx+E=x^2 \\ \implies Cx^2+(2C+D)x+2C+D+E=x^2 \\ \text{By comparison of the coefficients of } x^2 \text{ , x and the constant terms wehave } \\C=1 ;2C+D=0 \implies 2(1)+D=0\implies D=-2\\2C+D+E=0 \implies 2(1)-2+E=0 \implies E=0\\ \text{Then the particular integral is } Y_p=2x^2-2x\\ \text{Then the general solution of the D.E is given by: }\\ y(x)=Y_c+Y_p= e^{\frac{-x}{2}}(A\cos(\frac{3x}{2})+B\sin(\frac{3x}{2}))+2x^2-2x$