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In each of the following, $Q .1$ to 5 form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b$.
$y=e^x(a \cos x+b \sin x)$
$y=e^x(a \cos x+b \sin x)$
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Verified Answer
The curve is $y=e^x(a \cos x+b \sin x)$
Differentiating w.r.t. $x$
$$
\begin{aligned}
&y^{\prime}=e^x(a \cos x+b \sin x)+e^x(-a \sin x+b \cos x) \\
&y^{\prime}=e^x[(a+b) \cos x-(a-b) \sin x]
\end{aligned}
$$
$$
\begin{aligned}
&\left.y^{\prime \prime}=e^x[a+b) \cos x-(a-b) \sin x\right] \quad+e^x[-(a+b) \\
&=\quad \sin x-(a-b) \cos x]
\end{aligned}
$$
Adding (ii) and (iii)
$$
y+\frac{y^{\prime \prime}}{2}=e^x[(a+b) \cos x-(a-b) \sin x]=y^{\prime}
$$
or $2 \mathrm{y}+\mathrm{y}^{\prime \prime}=2 \mathrm{y}^{\prime} \quad \Rightarrow \mathrm{y}^{\prime \prime}-2 \mathrm{y}^{\prime}+2 \mathrm{y}=0$
Hence the diff. equ. is $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}-2 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0$
Differentiating w.r.t. $x$
$$
\begin{aligned}
&y^{\prime}=e^x(a \cos x+b \sin x)+e^x(-a \sin x+b \cos x) \\
&y^{\prime}=e^x[(a+b) \cos x-(a-b) \sin x]
\end{aligned}
$$
$$
\begin{aligned}
&\left.y^{\prime \prime}=e^x[a+b) \cos x-(a-b) \sin x\right] \quad+e^x[-(a+b) \\
&=\quad \sin x-(a-b) \cos x]
\end{aligned}
$$
Adding (ii) and (iii)
$$
y+\frac{y^{\prime \prime}}{2}=e^x[(a+b) \cos x-(a-b) \sin x]=y^{\prime}
$$
or $2 \mathrm{y}+\mathrm{y}^{\prime \prime}=2 \mathrm{y}^{\prime} \quad \Rightarrow \mathrm{y}^{\prime \prime}-2 \mathrm{y}^{\prime}+2 \mathrm{y}=0$
Hence the diff. equ. is $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}-2 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0$
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