Question

Given equation $Y - y' - 2y = 3x - 2$. How about \[{y_p}\] and \[{y_c}\].(\[{y_c}\]=? And \[{y_p}\]=?)

Answer 1

Hint: In this question we have to solve the differential equation using undetermined coefficients, first solve the characteristic equation of this differential, and use the fact that homogeneous part of solution is \[{y_c} = {C_1}{e^{{y_1}x}} + {C_2}{e^{{y_2}x}}\] and the particular solution must be in form:\[{y_p} = Ax + B\], now deriving the \[{y_p}\] two times and comparing this with the given equation we will get the coefficients, and by adding the equations \[{y_c}\] and \[{y_p}\], we will get the required answer.

Complete step by step answer:
Given equation is $Y - y' - 2y = 3x - 2$,
This can be written as,
$y'' - y' - 2y = 3x - 2$,
First the characteristic equation of this differential one will be,
\[ \Rightarrow {y^2} - y - 2 = 0\],
Now taking common term we get,
\[ \Rightarrow \left( {y - 2} \right)\left( {y + 1} \right) = 0\],
Hence the roots are,
\[ \Rightarrow \]\[{y_1} = 2\], and \[{y_2} = - 1\],
We know that homogeneous part of solution is \[{y_c} = {C_1}{e^{2x}} + {C_2}{e^{ - x}}\], and the particular solution must be in form: \[{y_p} = Ax + B\],
Now differentiating both sides we get,
\[ \Rightarrow y{'_p} = A\],
Now again differentiating on both sides we get,
\[ \Rightarrow y'{'_p} = 0\],
From the above we get, by substituting the values in the given equation we get,
\[ \Rightarrow \]\[y'' - y' - 2y = 0 - A - 2\left( {Ax + B} \right)\],
Now simplifying we get,
\[ \Rightarrow \]\[y'' - y' - 2y = - A - 2Ax - 2B\],
Now combining the like terms we get,
\[ \Rightarrow y'' - y' - 2y = - \left( {A + 2B} \right) - 2Ax\],
Now we know that $Y - y' - 2y = 3x - 2$, now comparing the terms we get,
\[ \Rightarrow - \left( {A + 2B} \right) - 2Ax = 3x - 2\],
Now comparing we get,
\[ \Rightarrow - 2A = 3\], and \[ - \left( {A + 2B} \right) = - 2\],
Now simplifying we get,
\[ \Rightarrow A = \dfrac{{ - 3}}{2}\],----(1)
Now substituting the value in (1) we get,
\[ \Rightarrow \dfrac{{ - 3}}{2} + 2B = 2\],
Now simplifying we get,
\[ \Rightarrow 2B = 2 + \dfrac{3}{2}\],
Now adding we get,
\[ \Rightarrow 2B = \dfrac{7}{2}\],
Now dividing 2 on both sides we get,
\[ \Rightarrow \dfrac{{2B}}{2} = \dfrac{{\dfrac{7}{2}}}{2}\],
Now simplifying we get,
\[ \Rightarrow B = \dfrac{7}{4}\],
Now substituting the values in the equation \[{y_p} = Ax + B\], we get,
\[ \Rightarrow {y_p} = \dfrac{{ - 3}}{2}x + \dfrac{7}{4}\],
So, finally \[{y_p} = \dfrac{{ - 3}}{2}x + \dfrac{7}{4}\] and \[{y_c} = {C_1}{e^{2x}} + {C_2}{e^{ - x}}\].

\[\therefore \] By solving differential equation $Y - y' - 2y = 3x - 2$ by the value of \[{y_p}\] and \[{y_c}\] will be \[{y_p} = \dfrac{{ - 3}}{2}x + \dfrac{7}{4}\] and \[{y_c} = {C_1}{e^{2x}} + {C_2}{e^{ - x}}\].

Note: The method of finding a particular solution \[{y_p}\], first make an assumption about the form of \[{y_p}\], then find the coefficients, and general rule for assumption is that linear combination of all the linearly independent types of functions that aside from repeated differentiations. No function in the assumed \[{y_p}\] duplicates any part of \[{y_c}\].