Question

In Problems 35–64 solve the given differential equation by undetermined coefficients.

$\mathbf{2}\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{7}\mathbf{y}\text{'}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{29}$

Short answer

$y=C_1{e}^x+C_2{e}^{\frac{5}{2}x}-\frac{29}{5}$

Step-by-step solution

Definition of homogeneous differential equation

A homogeneous differential equation is an equation containing a differentiation and a function, with a set of variables.

Find the complementary function

From$2m^2-7m+5=0$we find$m_1=\frac{5}{2}$and$m_2=1$

So, the complementary function is$y_c=C_1{e}^x+C_2{e}^{\frac{5x}{2}}$.

Now, since$D(-29)=0$, we apply the differential operatorto both sides of given equation, to get

$$\begin{gathered} D\left(2D^2-7D+5\right)=D(-29) \\ =0 \end{gathered}$$

Evaluation

The auxiliary equation of the above equation is

$$\begin{gathered} m\left(2m^2-7m+5\right)=0 \\ \Rightarrow m_1=\frac{5}{2},m_2=1,m_3=0 \end{gathered}$$

So, the general solution must be $y=C_1{e}^x+C_2{e}^{\frac{5}{2}x}+C_3$.

Substitution

Since$y=y_c+y_p$, and noticing that the first two terms are already present in the complementary function, the particular solution should be of the following form:

$y_p=A$

Now, substituting the above into the given differential equation yields

$$\begin{gathered} 5A=-29 \\ A=-\frac{29}{5} \end{gathered}$$

Thus, the general solution is$y=C_1{e}^x+C_2{e}^{\frac{5}{2}x}-\frac{29}{5}$.