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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 $b$.
$y=a e^{3 x}+b e^{-2 x}$
$y=a e^{3 x}+b e^{-2 x}$
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$\mathrm{y}=\mathrm{ae} \mathrm{e}^{3 \mathrm{x}}+$ be $^{-2 \mathrm{x}}$
Differentiating w.r.t. $\mathrm{x}: \mathrm{y}^{\prime}=3 \mathrm{ae} \mathrm{e}^{3 \mathrm{x}}-2 \mathrm{be}-2 \mathrm{x}$
Again differentiating: $\mathrm{y}^{\prime \prime}=9 a \mathrm{ac}^{3 \mathrm{x}}+4 \mathrm{be}^{-2 \mathrm{x}}$
Multiply equation (i) by 2 and add with (ii) we get
i.e., $2 \mathrm{y}=2 \mathrm{a} \mathrm{e}^{3 \mathrm{x}}+2 \mathrm{be} \mathrm{e}^{-2 \mathrm{x}}$
$$
\frac{\mathrm{y}^{\prime}=3 \mathrm{a} \mathrm{e}^{3 \mathrm{x}}-2 \mathrm{be}^{-2 \mathrm{x}}}{2 \mathrm{y}+\mathrm{y}^{\prime}=5 \mathrm{a} \mathrm{e}^{3 \mathrm{x}}} \text { or } a \mathrm{e}^{3 \mathrm{x}}=\frac{2 \mathrm{y}+\mathrm{y}^{\prime}}{5}
$$
Multiply (i) by 3 and subtract (ii) from $3 \mathrm{y}-\mathrm{y}^{\prime}=5 \mathrm{be}^{-2 \mathrm{x}} \quad$ or $\quad$ be $\mathrm{e}^{-2 \mathrm{x}}=\frac{3 \mathrm{y}-\mathrm{y}^{\prime}}{5}$
Putting the values of a and b in (iii), we get; $5 y^{\prime \prime}=30 y+5 y^{\prime} \quad$ or $\quad \frac{d^2 y}{d x^2}-\frac{d y}{d x}-6 y=0$
Which is the req. differential equation
Differentiating w.r.t. $\mathrm{x}: \mathrm{y}^{\prime}=3 \mathrm{ae} \mathrm{e}^{3 \mathrm{x}}-2 \mathrm{be}-2 \mathrm{x}$
Again differentiating: $\mathrm{y}^{\prime \prime}=9 a \mathrm{ac}^{3 \mathrm{x}}+4 \mathrm{be}^{-2 \mathrm{x}}$
Multiply equation (i) by 2 and add with (ii) we get
i.e., $2 \mathrm{y}=2 \mathrm{a} \mathrm{e}^{3 \mathrm{x}}+2 \mathrm{be} \mathrm{e}^{-2 \mathrm{x}}$
$$
\frac{\mathrm{y}^{\prime}=3 \mathrm{a} \mathrm{e}^{3 \mathrm{x}}-2 \mathrm{be}^{-2 \mathrm{x}}}{2 \mathrm{y}+\mathrm{y}^{\prime}=5 \mathrm{a} \mathrm{e}^{3 \mathrm{x}}} \text { or } a \mathrm{e}^{3 \mathrm{x}}=\frac{2 \mathrm{y}+\mathrm{y}^{\prime}}{5}
$$
Multiply (i) by 3 and subtract (ii) from $3 \mathrm{y}-\mathrm{y}^{\prime}=5 \mathrm{be}^{-2 \mathrm{x}} \quad$ or $\quad$ be $\mathrm{e}^{-2 \mathrm{x}}=\frac{3 \mathrm{y}-\mathrm{y}^{\prime}}{5}$
Putting the values of a and b in (iii), we get; $5 y^{\prime \prime}=30 y+5 y^{\prime} \quad$ or $\quad \frac{d^2 y}{d x^2}-\frac{d y}{d x}-6 y=0$
Which is the req. differential equation
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