Differential Equations — Class 12 Maths Solution

Given that, $\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + y = 0$

${D^2}y - 2Dy + y = 0$

where $D = \frac{d}{{dx}}$

$\left( {{D^2} - 2D + 1} \right)y = 0$

The auxiliary equation is
${m^2} - 2m + 1 = 0$
${(m - 1)^2} = 0 \Rightarrow m = 1,1$

Since, the roots are real and equal.

$\therefore$ ${\rm{CF}} = (Ax + B){e^x} \Rightarrow y = (Ax + B){e^x}$

[since, if roots of Auxilliary equation are real and equal say $(m)$,

then ${\rm{CF}} = \left( {{C_1}x + {C_2}} \right){e^{mx}}$]