Question

Solve the following differential equation:
$\left( {{D^2} - 6D + 9} \right)y = x + {e^{2x}}$

Answer 1

In this particular question use the concept that the solution of any differential equation is the sum of complementary function and particular integral, so find the complementary function by equating the L.H.S of the equation to zero and find it roots, then the general solution of the complementary function if both the roots are same is written as, $\left( {{C_1} + {C_2}x} \right){e^{mx}}$, where m is the same roots so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given differential equation
$\left( {{D^2} - 6D + 9} \right)y = x + {e^{2x}}$
Now as we know that the solution of the above equation is the sum of complementary function and particular integral.
Therefore, y = C.F + P.I
Now first find out complementary functions.
So for this equation L.H.S of the above equation to zero so we have,
$\Rightarrow \left( {{D^2} - 6D + 9} \right)y = 0$
$\Rightarrow \left( {{D^2} - 6D + 9} \right) = 0$
Now factorize the above equation we have,
$\Rightarrow \left( {{D^2} - 3D - 3D + 9} \right) = 0$
$\Rightarrow \left( {D - 3} \right)\left( {D - 3} \right) = 0$
$\Rightarrow D = 3,3$
So the solution of the complementary function is given as,
C.F = $\left( {{C_1} + {C_2}x} \right){e^{3x}}$, where ${C_1}{\text{ and }}{C_2}$ are some arbitrary constants.
Therefore, C.F = $\left( {{C_1} + {C_2}x} \right){e^{3x}}$
Now for Particular integral we have,
$\Rightarrow \left( {{D^2} - 6D + 9} \right)y = x + {e^{2x}}$
$\Rightarrow y = \dfrac{{x + {e^{2x}}}}{{{D^2} - 6D + 9}}$
The above equation is also written as
$\Rightarrow y = \dfrac{x}{{{D^2} - 6D + 9}} + \dfrac{{{e^{2x}}}}{{{D^2} - 6D + 9}}$
$\Rightarrow y = \dfrac{x}{{{{\left( {D - 3} \right)}^2}}} + \dfrac{{{e^{2x}}}}{{{D^2} - 6D + 9}}$
$\Rightarrow y = \dfrac{x}{{{{\left( { - 3} \right)}^2}{{\left( {1 - \dfrac{D}{3}} \right)}^2}}} + \dfrac{{{e^{2x}}}}{{{D^2} - 6D + 9}}$
$\Rightarrow y = \dfrac{x}{9}{\left( {1 - \dfrac{D}{3}} \right)^{ - 2}} + \dfrac{{{e^{2x}}}}{{{D^2} - 6D + 9}}$
Now according to the binomial expansion the expansion of ${\left( {1 - D} \right)^{ - n}} = 1 + nD$
We neglect the higher terms of D as in the above expression, outside ${\left( {1 - \dfrac{D}{3}} \right)^{ - 2}}$ this expression there is only x and D is the symbol of the differentiation, so higher order differentiation of x is zero, so use this property in the above expression we have,
$\Rightarrow y = \dfrac{x}{9}\left( {1 + 2\dfrac{D}{3}} \right) + \dfrac{{{e^{2x}}}}{{{D^2} - 6D + 9}}$
$\Rightarrow y = \dfrac{x}{9} + \dfrac{1}{9}.\dfrac{2}{3}\left( {\dfrac{d}{{dx}}x} \right) + \dfrac{{{e^{2x}}}}{{{D^2} - 6D + 9}}$
Now in the above equation the coefficient of x in the power of exponential is 2, so substitute in place of D= 2 in the second expression of the above equation, so the particular integral is.
$\Rightarrow P.I = \dfrac{x}{9} + \dfrac{1}{9}.\dfrac{2}{3}\left( 1 \right) + \dfrac{{{e^{2x}}}}{{{2^2} - 6\left( 2 \right) + 9}}$, $\left[ {\because \dfrac{d}{{dx}}x = 1} \right]$
$\Rightarrow P.I = \dfrac{x}{9} + \dfrac{2}{{27}} + \dfrac{{{e^{2x}}}}{{13 - 12}}$
$\Rightarrow P.I = \dfrac{x}{9} + \dfrac{2}{{27}} + {e^{2x}}$
So the solution of the given differential equation is,
$\Rightarrow y = \left( {{C_1} + {C_2}x} \right){e^{3x}} + \dfrac{x}{9} + \dfrac{2}{{27}} + {e^{2x}}$
So this is the required solution of the given differential equation.

Note Whenever we face such types of questions the key concept we have to remember is that always recall that there are multiple methods to solve the particular integral depending on the R.H.S of the differential equation, if the R.H.S is in the form of ${e^{px}}$, then substitute in place of D i.e.$\dfrac{d(px)}{dx}$, p and solve as above then add these two solutions as above we will get the required answer.