Question

If $y=e^{4x}+2e^{-x}$ satisfies the differential equation $y_3+A{y}_1+By=0$, then

• A. $A=-13,B=-12$
• B. $A=13,B=12$
• C. $A=-12,B=-13$
• D. $A=12,B=13$

Answer 1

The correct option is A $A=-13,B=-12$

$y=e{}^{4x}+2e^{-x}$ (given)

Differentiating both sides
$y_1=4e{}^{4x}-2e{}^{-x}$

Again differentiating

$y_2=16e{}^{4x}+2e^{-x}$

Again differentiating, we get
$y_3=64e{}^{4x}-2e{}^{-x}$

Substituting these derivatives in
$y_3+A{y}_1+By=0$

$\Rightarrow 64e{}^{4x}-2e{}^{-x}+4Ae{}^{4x}-2Ae{}^{-x}+Be{}^{4x}+2Be{}^{-x}=0$

$\Rightarrow e{}^{4x}(64+4A+B)+e{}^{-x}(-2-2A+2B)=0$

Only possibility when the term in parenthesis is zero,

$\Rightarrow 4A+B=-64$ and $2B-2A=2$

$A=-13$

$B=-12$

Hence, option A.