Question
If y = 500e7x + 600e-7x show that $\frac{{{d^2}y}}{{d{x^2}}} = 49y$.
$\therefore \frac{{dy}}{{dx}} $ = 500e7x(7) + 600e-7x(-7) = 500(7)e7x - 600(7)e-7x
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} $ = 500(7)e7x(7) - 600(7)e-7x(-7) = 500(49)e7x + 600(49){e-7x
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} $ = 49[500e7x + 600e-7x]
= 49y [From eq. (i)]
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 49y$ Hence proved.
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f(x) = 2x3 + 5 on R.
$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$ and, $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}+\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$