You visited us 1 times! Enjoying our articles? Unlock Full Access!

Homogeneous Linear Differential Equations (General Form of LDE)

The maximum v...

Question

The maximum value of the solution $y (t)$ of the differential equation $y (t) + . . y (t) = 0$ with the initial conditions $. y (0) = 1$ and $y (0) = 1,$ for $t \geq 0$ is

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\sqrt{2}$
Given $. . y (t) + y (t) = 0$
$\Rightarrow$ $(D^{2} + 1) y = 0$ ... $(1)$
A.E is $m^{2} + 1 = 0 \Rightarrow m = \pm i$
The general solution of equation $(1)$ is
$y = C_{1} c o s t + C_{2} s i n t$
Using $y (0) = 1$ & $. y (0) = 1$ , we get $C_{1} = 1$ and $C_{2} = 1$
$∴$ The solution of equation $(1)$ is
$y = c o s t + s i n t$ ... $(2)$
For maximum (or) minimum of equation $(2)$ , we have
$\frac{d y}{d t} = 0$
$- s i n t + c o s t = 0$ or $t a n t = 1 \Rightarrow t = \frac{π}{4}$
Hence, the maximum value of equation $(2)$ is
$y (\frac{π}{4}) = c o s (\frac{π}{4}) + s i n (\frac{π}{4}) = \sqrt{2}$

Suggest Corrections

Similar questions

y (t)

y (0) = 1

y (1) = 3 e^{- 1}

\frac{d^{2} y}{d t^{2}} + 2 \frac{d y}{d t} + y = 0

y (2)

y (t)

(1 + t) \frac{d y}{d t} - t y = 1

y (0) = - 1

y (1)

y (t)

t > 0

x (t)

y (0)

- 2 y (t)

t > 0

\frac{d^{2} y (t)}{d t^{2}} + 2 \frac{d y (t)}{d t} + y (t) = δ (t) w i t h y (t) |_{t = 0} = - 2 a n d \frac{d y}{d t} |_{t = 0} = 0.

\frac{d y}{d t} |_{t = 0}

y (t)

(1 + t) \frac{d y}{d t} - t y = 1

y (0) = - 1

y (1)

Join BYJU'S Learning Program