Question

Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection or by$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{18}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{23}\mathbf{)}\mathbf{,}$ or. $\mathbf{(}\mathbf{6}\mathbf{.24}\mathbf{)}$Alsofind a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form [see$\mathbf{(}\mathbf{6}\mathbf{.26}\mathbf{)}$ andthe discussion after]$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{15}\mathbf{)}$.

$\mathbf{5}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathit{x}$

Short answer

The general solution given by differential equation is$y=e^{-3x/5}({\alpha }_1\sin \frac{x}{5}+{\alpha }_2\cos \frac{x}{5})+\frac{x^2-5}{2}$

Step-by-step solution

Given data. 

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

General solution of differential equation

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.

Find the general solution of given differential equation.5y''+6y'+2y=x2+6x

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,

$y_c=e^{-3x/5}({\alpha }_1\sin \frac{x}{5}+{\alpha }_2\cos \frac{x}{5})$

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

$y=y_c+y_p=e^{-3x/5}({\alpha }_1\sin \frac{x}{5}+{\alpha }_2\cos \frac{x}{5})+\frac{x^2-5}{2}$