Question

A differential equation associated to the primitive $y=a+b{e}^{5x}+c{e}^{-7x}$ is (where $y_n$ is the nth derivative w.r.t.$x$)

• A. ${\mathrm{y}}_3+3{\mathrm{y}}_2–{\mathrm{y}}_1=0$
• B. $4{\mathrm{y}}_3+5{\mathrm{y}}_2–20{\mathrm{y}}_1=0$
• C. ${\mathrm{y}}_3+2{\mathrm{y}}_2–35{\mathrm{y}}_1=0$
• D. none of these

Answer 1

The correct option is C

${\mathrm{y}}_3+2{\mathrm{y}}_2–35{\mathrm{y}}_1=0$

Explanation for the correct option.

$y=a+b{e}^{5x}+c{e}^{-7x}$

Differentiating w.r.t $x$

$y_1=5b{e}^{5x}–7c{e}^{-7x}...\left(i\right)$

$y_2=25b{e}^{5x}+49c{e}^{-7x}..\left(ii\right)$

$y_3=125b{e}^{5x}–343c{e}^{-7x}..\left(iii\right)$

Multiply (i) by $7$

$7y_1=35b{e}^{5x}–49c{e}^{-7x}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{iv}\mathbf{\right)}$

Add (ii) and (iv)

$y_2+7y_1=60b{e}^{5x}..\left(v\right)$

Multiply (i) by $5$

$5y_1=25b{e}^{5x}–35c{e}^{-7x}..\left(vi\right)$

Subtract (vi) from (ii)

$y_2-5y_1=84c{e}^{-7x}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{vii}\mathbf{\right)}$

Putting the values of $b{e}^{5x}$ and $c{e}^{-7x}$ in (iii) and solving we get

$y_3+2y_2-35y_1=0$

Hence, option (C) is the answer.