Q119 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Mathematics

Statement (I) The elimination of arbitrary constants from \(\alpha, \beta\) and \(\gamma\) from \(y=(\alpha+\beta+\gamma) x\) results in a differential equation of order three. Statement (II) The elimination of arbitrary constants \(\alpha, \beta\) and \(\gamma\) from \(y=\alpha x+\beta \sin x+\gamma e^x\) results in a differential equation of order three.

  • A
    I is true and II is false
  • B
    I is false and II is false
  • C
    I is true and II is true
  • Correct answer
    D
    I is false and II is true

Solution

Statement I \(\Rightarrow \quad y=(\alpha+\beta+\gamma) x\) Differentiate both sides w.r.t. \(x\), we get \(\frac{d y}{d x}=(\alpha+\beta+\gamma)=k\) Here, order \(=1\) So, statement \(\mathrm{I}\) is false. Statements II : \(y=\alpha x+\beta \sin x+\gamma e^x\) Differentiate both sides w.r.t. \(x\), we get \(\frac{d y}{d x}=\alpha+\beta \cos x+\gamma e^x\)...(i) Again, differentiating both sides w. r. t. \(x\), we get \(\Rightarrow \quad \begin{aligned} \frac{d^2 y}{d x^2} & =-\beta \sin x+\gamma e^x \quad \ldots (ii) \\ \Rightarrow \quad \frac{d^3 y}{d x^2} & =-\beta \cos x+\gamma e^x \quad \ldots (iii) \end{aligned}\) Subtracting Eq. (ii) from Eq. (iii), we get \(\begin{aligned} \frac{d^3 y}{d x^2}-\frac{d^2 y}{d x^2} & =-\beta \cos x+\gamma e^x+\beta \sin x-\gamma e^x \\ & =\beta \sin x-\beta \cos x \\ & =\beta(\sin x-\cos x) \end{aligned}\) Hence, order \(=3\) So, statement II is true.

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

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