Question

The solution of $\frac{d^{3}y}{d{x}^3}-4\frac{d^{2}y}{d{x}^2}+5\frac{dy}{dx}-2y=0$ is:

• A. $y=c_1{e}^x+c_2{e}^{2x}+c_3{e}^{3x}$
• B. $y=(c_1+c_2+c_3{x}^2)e^x$
• C. $y=(c_1+c_{2}x)e^x+c_3{e}^{2x}$
• D. $y=(c_1+c_2{x}^2)e^x+c_3{e}^{2x}$

Answer 1

The correct option is C $y=(c_1+c_{2}x)e^x+c_3{e}^{2x}$

The given equation can be wriitten as
$(\frac{d}{dx}-1)(\frac{d}{dx}-1)(\frac{d}{dx}-2)y=0$
Let $(\frac{d}{dx}-1)(\frac{d}{dx}-2)y=0$
Then $\frac{dv}{dx}-u=0\Rightarrow u=c_1^{\prime }{e}^x$
Therefore, we have
$(\frac{d}{dx}-1)(\frac{dy}{dx}-2y)=c_1^{\prime }{e}^x$
Putting $v=\frac{dy}{dx}-2y$
We have $\frac{dv}{dx}-u=c_1^{\prime }{e}^x$
whose I.F is $e^{-x}$ so
$v{e}^{-x}=c_1^{\prime }x+c_2^{\prime }\Rightarrow v=(c_1^{\prime }x+c_2^{\prime })e^x$
Hence $\frac{dy}{dx}-2y=(c_1^{\prime }x+c_2^{\prime })e^x$
which is again a linear equation with I.F $e^{-2x}$
Hence
$y{e}^{-2x}=\int (c_1^{\prime }x+c_2^{\prime })e^{-x}dx+c_3=-(c_1^{\prime }x+c_2^{\prime })e^{-x}-(c_1^{\prime }x+c_2^{\prime })e^{-x}-c_1^{\prime }x{e}^{-x}+c_3\Rightarrow y=(c_{1}x+c_2)e^x+e^{2x}{c}_3$