Question

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

$\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{3}\mathbf{y}\text{'}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{5}$

Short answer

$y=C_1+C_2{e}^{-3x}-\frac{19}{9}x+\frac{2}{3}{x}^2$

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.

Given data

From$m^2+3m=0$we find$m_1=0$and$m_2=-3$

So, the complementary function is $y_c=C_1+C_2{e}^{-3x}$

Apply differential operator

Since$D^2(4x-5)=0$, We apply the differential operator$D^2$to the both sides of the given equation to get

$\begin{array}{rcl}{D}^2\left(D^2+3D\right)y& =& D^2(4x-5)\\ & =& 0\end{array}$

Now, the auxiliary equation is

$$\begin{gathered} m^2\left(m^2+3m\right)=0 \\ \Rightarrow m_1=-3,m_2=0=m_3=m_4 \end{gathered}$$

Thus, the general solution must be .$y=\underset{y_c}{\underbrace{C_1+C_2{e}^{-3x}}}+C_{3}x+C_4{x}^2$

Substitution

Now, particular solution,

So, the terms remaining in y must be the basic form of $y_p$;

So, $y_p=Ax+B{x}^2$

Then

$\begin{array}{rcl}{\dot{y}}_p\text{'}& =& A+2Bx\\ & =& 2B\end{array}$

Now substituting these into the given differential equation gives

$$\begin{gathered} 2B+3(A+2Bx)=4x-5 \\ 6Bx+(2B+3A)=4x-5 \end{gathered}$$

Equating co-efficient

On equating the coefficients of like terms, we get

$\begin{array}{rcl}6B& =& 4\\ 2B+3A& =& -5\\ B& =& \frac{2}{3},A=-\frac{19}{9}\end{array}$

So, $y_p=-\frac{19}{9}x+\frac{2}{3}{x}^2$

Thus, the general solution is $y=C_1+C_2{e}^{-3x}-\frac{19}{9}x+\frac{2}{3}{x}^2$