91Ó°ÊÓ

Given equation is$5y^{\text{'}\text{'}}+6y^{\text{'}}+2y=x^2+6x$

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

It is clear that the right hand side of this differential equation is a polynomial of degree 2 , we know that the particular solution would be in the form$y_p=A{x}^2+Bx+C$, where$A,B$and$C$are undetermined coefficients which satisfy the differential equation we have that we are going to find. First we need to differentiate the particular solution two times, that is

$\begin{array}{l}{y}_p^{\text{'}}=2Ax+B\\ y_p^{\text{'}\text{'}}=2A\end{array}$

then we substitute in the differential equation we have

$\begin{array}{r}10A+12Ax+6B+2A{x}^2+2Bx+2C=x^2+6x\\ \left(2A\right)x^2+(12A+2B)x+(10A+6B+2C)=x^2+6x\end{array}$

Now, by comparing the corresponding coefficients we can find that,$A=\frac{1}{2},B=0$and.$C=−\frac{5}{2}$Therefore, the particular solution is

$y_p=\frac{x^2}{2}−\frac{5}{2}$

Next, we want to find the complementary solution, and to do so let us write the auxiliary equation

$\begin{array}{r}(5D^2+6D+2)y=x^2+6x\\ (D+\frac{3−i}{5})(D+\frac{3+i}{5})y=x^2+6x\end{array}$

Then we need to solve it but when the right-hand side of it is zero, that is

$(D+\frac{3−i}{5})(D+\frac{3+i}{5})y=0$

The roots of the auxiliary equation are complex and not equal, that is,

$y_c=e^{−3x/5}({\mathrm{Î\pm }}_1\sin \frac{x}{5}+{\mathrm{Î\pm }}_2\cos \frac{x}{5})$

Now we can write the general solution of the differential equation (which is the linear sum of the complementary solution and the particular solution),

$y=y_c+y_p=e^{−3x/5}({\mathrm{Î\pm }}_1\sin \frac{x}{5}+{\mathrm{Î\pm }}_2\cos \frac{x}{5})+\frac{x^2−5}{2}$