Question

The differential equation of the family of curves $\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}(\mathrm{a}\cos \mathrm{x}+\mathrm{b}\sin \mathrm{x})$, where $a$ and $b$ are arbitrary constants, is given by

• A. ${\mathrm{y}}_2–4{\mathrm{y}}_1+5\mathrm{y}=0$
• B. $2{\mathrm{y}}_2–{\mathrm{y}}_1+5\mathrm{y}=0$
• C. ${\mathrm{y}}_2+4{\mathrm{y}}_1–5\mathrm{y}=0$
• D. ${\mathrm{y}}_2–2{\mathrm{y}}_1+5\mathrm{y}=0$

Answer 1

The correct option is A

${\mathrm{y}}_2–4{\mathrm{y}}_1+5\mathrm{y}=0$

Explanation for the correct answer:

Given curve,

$$\begin{gathered} \mathrm{y}={\mathrm{e}}^{2\mathrm{x}}(\mathrm{a}\cos \mathrm{x}+\mathrm{b}\sin \mathrm{x}) \\ {\mathrm{y}}_1=\frac{d\mathrm{y}}{d\mathrm{x}} \\ {\mathrm{y}}_2=\frac{d{\mathrm{y}}_1}{d\mathrm{x}} \end{gathered}$$

On Differentiating, we get

$$\begin{gathered} \frac{d\mathrm{y}}{d\mathrm{x}}={\mathrm{e}}^{2\mathrm{x}}·2(\mathrm{a}\cos \mathrm{x}+\mathrm{b}\sin \mathrm{x})+{\mathrm{e}}^{2\mathrm{x}}(-\mathrm{a}\sin \mathrm{x}+\mathrm{b}\cos \mathrm{x}) \\ {\mathrm{y}}_1=2\mathrm{y}+{\mathrm{e}}^{2\mathrm{x}}(-\mathrm{a}\sin \mathrm{x}+\mathrm{b}\cos \mathrm{x})...\left(1\right) \\ \frac{d{\mathrm{y}}_1}{d\mathrm{x}}=\frac{2d\mathrm{y}}{d\mathrm{x}}+2{\mathrm{e}}^{2\mathrm{x}}(-\mathrm{a}\sin \mathrm{x}+\mathrm{b}\cos \mathrm{x})+{\mathrm{e}}^{2\mathrm{x}}(-\mathrm{a}\cos \mathrm{x}-\mathrm{b}\sin \mathrm{x}) \end{gathered}$$

$$\begin{gathered} {\mathrm{y}}_2=2{\mathrm{y}}_1+2{\mathrm{e}}^{2\mathrm{x}}\left(-a\mathrm{sinx}+b\mathrm{cosx}\right)-\mathrm{y} \\ {\mathrm{y}}_2=-\mathrm{y}+2\left({\mathrm{y}}_1-2\mathrm{y}\right)+2{\mathrm{y}}_1[\mathrm{From}(1\left)\right] \\ {\mathrm{y}}_2=-\mathrm{y}+4{\mathrm{y}}_1-4\mathrm{y} \end{gathered}$$

$\Rightarrow {\mathrm{y}}_2-4{\mathrm{y}}_1+5\mathrm{y}=0$

Hence, the correct option is (A)