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Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)2 Marks
Question
Obtain the differential equation by eliminating arbitrary constants from the following equations : $y=A e^{3 x}+B e^{-3 x}$
✓

Answer

$
y=A e^{3 x}+B e^{-3 x}
$
Differentiating twice w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d y}{d x}=A e^{3 x} \times 3+B e^{-3 x} \times(-3) \\
& \begin{aligned}
\therefore \frac{d y}{d x}= & 3 A e^{3 x}-3 B e^{-3 x} \\
\text { and } \frac{d^2 y}{d x^2} & =3 A e^{3 x} \times 3-3 B e^{-3 x} \times(-3) \\
& =9 A e^{3 x}+9 B e^{-3 x} \\
& =9\left(A e^{3 x}+B e^{-3 x}\right)=9 y
\end{aligned} \\
& \begin{aligned}
\therefore \frac{d^2 y}{d x^2} & =9 y
\end{aligned}
\end{aligned}
$
This is the required D.E.

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