Q118 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics

The differential equation corresponding to the family of curves given by $y=a+b e^{2 x}+c e^{-3 x}$ is

  • A
    $y_3-y_2+6 y_1=0$
  • Correct answer
    B
    $y_3+y_2-6 y_1=0$
  • C
    $y_3-6 y_2-y_1=0$
  • D
    $y_3+6 y_2-y_1=0$

Solution

We have, $$ y=a+b e^{2 x}+c e^{-3 x} $$ Differentiate w.r.t. ' $x$ ' $$ y_1=\frac{d y}{d x}=2 b e^{2 x}-3 c e^{-3 x} $$ Again, differentiate w.r.t. ' $x$ ' $$ y_2=\frac{d^2 y}{d x^2}=4 b e^{2 x}+9 c e^{-3 x} $$ Again, differentiate w.r.t. ' $x$ ' $$ y_3=\frac{d^3 y}{d x^3}=8 b e^{2 x}-27 c e^{-3 x} $$ Now adding Eqs. (iii) and (iv) and subtracting $6 \times$ Eq. (ii), $$ \begin{aligned} & y_2+y_3-6 y_1=\left(4 b e^{2 x}+9 c e^{-3 x}\right) \\ & +\left(8 b e^{2 x}-27 c e^{-3 x}\right)-6\left(2 b e^{2 x}-3 c e^{-3 x}\right) \\ = & \left(12 b e^{2 x}-18 c e^{-3 x}\right)-\left(12 b e^{2 x}-18 c e^{-3 x}\right)=0 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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