Question

d³y/dx³+2 d²y/dx²+dy/dx=e^-2x+sin 2x​

Answer 1

The general solution to the differential equation is

$y = (A + Bx)e^{-x} + C + \dfrac{1}{50} [-25e^{-2x}-4sin2x+3cos2x]$

Step-by-step explanation:

Given: Differential Equation $\frac{d^{3}y }{dx^{3} } + 2 \frac{d^{2}y }{dx^{2} } + \frac{dy }{dx} = e^{-2x}+sin 2x$

To Find: General solution of the given differential equation

Solution:

• Finding the general solution of the given differential equation

The given differential equation can be written in the following way,

$\Rightarrow (D^{3} + 2D^{2} + D)y = e^{-2x}+sin 2x$

$\Rightarrow D ( D^{2} + 2D + 1)y = e^{-2x}+sin 2x$

where D=d/dx

To find the complementary factor, consider the above differential equation such that,

$\Rightarrow m ( m^{2} + 2m + 1)= 0$

$\Rightarrow m ( m + 1 )^{2} = 0$

$\Rightarrow m = 0, -1, -1$

Therefore, $y_{CF} = Ce^{(0)x} + (A + Bx)e^{-x}$

To find the particular integral, consider the above differential equation such that,

$y_{PI} = \dfrac{1}{D ( D^{2} + 2D + 1)} e^{-2x}+sin 2x$

$\Rightarrow y_{PI} = \dfrac{1}{D ( D^{2} + 2D + 1)} e^{-2x}+ \dfrac{1}{D ( D^{2} + 2D + 1)} sin 2x$

using the rules $\frac{1}{f(D)}e^{ax} = \frac{1}{f(a)}e^{ax}$ and $\frac{1}{f(D^2)}sinax = \frac{1}{f(-a^2)} sinax$ in the 1st and 2nd terms of the above equation respectively, we get,

$\Rightarrow y_{PI} = \dfrac{1}{(-2) ( (-2)^{2} + 2(-2) + 1)} e^{-2x}+ \dfrac{1}{D ( -(2)^{2} + 2D + 1)} sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x}+ \dfrac{1}{D ( 2D - 3)} sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x}+ \dfrac{1}{( 2D^2 - 3D)} sin 2x$

again using the rule $\frac{1}{f(D^2)}sinax = \frac{1}{f(-a^2)} sinax$ in the 2nd term, we get,

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x}+ \dfrac{1}{( 2(-(2)^2) - 3D)} sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \dfrac{1}{( 8 + 3D)} sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \Big[ \dfrac{1}{( 8 + 3D)} \times \dfrac{( 8 - 3D)}{( 8 - 3D)} \Big] sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \dfrac{( 8 - 3D)}{( 64 - 9D^2)} sin 2x$

again using the rule $\frac{1}{f(D^2)}sinax = \frac{1}{f(-a^2)} sinax$ in the 2nd term, we get,

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \dfrac{( 8 - 3D)}{( 64 - 9(-4))} sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \dfrac{( 8 - 3D)}{100} sin 2x$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \dfrac{4}{50} sin 2x + \dfrac{3D(sin2x)}{100}$

$\Rightarrow y_{PI} = -\dfrac{1}{2} e^{-2x} - \dfrac{4}{50} sin 2x + \dfrac{3(2cos2x)}{100}$

$\Rightarrow y_{PI} = \dfrac{1}{50} [- 25e^{-2x} - 4 sin 2x + 3cos2x]$

Therefore, the general solution is $y = y_{CF} + y_{PI} = (A + Bx)e^{-x} + C + \dfrac{1}{50} [-25e^{-2x}-4sin2x+3cos2x]$

Hence, the general solution to the differential equation is

$y = (A + Bx)e^{-x} + C + \dfrac{1}{50} [-25e^{-2x}-4sin2x+3cos2x]$