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

Byju's Answer

Other

Engineering Mathematics

Homogeneous Linear Differential Equations (General Form of LDE)

It is given t...

Question

It is given that $y^{''} + 2 y^{'} + y = 0, y (0) = 0, y (1) = 0$ . What is $y (0.5)$ ?

0

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

0.37

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

0.62

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

1.13

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

Open in App

Solution

The correct option is A $0$
Given the differential equation
$y^{2} + 2 y^{'} + y = 0$ (i)
Boundary conditions are
$y (0) = y (1) = 0$
Auxilary equation of (i)
$m^{2} + 2 m + 1 = 0$
$\Rightarrow m = - 1, - 1$
General soluiton of (i) is
$y (x) = (A + B x) e^{- x}$
Now: $y (0) = 0$
$\Rightarrow 0 = (A + B .0) e^{o}$
$\Rightarrow A = 0$
and $y (1) = 0$
$\Rightarrow 0 = (0 + B .1) e^{- 1}$
$\Rightarrow B = 0$
$y (x) = 0$
Hence, $y (0, 5) = 0$

Suggest Corrections

Similar questions

Q. (i)

\frac{d y}{d x} = y \tan x, y (0) = 1

(ii)

2 x \frac{d y}{d x} = 5 y, y (1) = 1

(iii)

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

(iv)

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

(v)

\frac{d y}{d x} = 2 x y, y (0) = 1

(vi)

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1

(vii)

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

(viii)

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}

when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)

2 (y + 3) - x y \frac{d y}{d x} = 0

, y(1) = −2 [NCERT EXEMPLAR]

Q. (i)

\frac{d y}{d x} = y \tan x, y (0) = 1

(ii)

2 x \frac{d y}{d x} = 5 y, y (1) = 1

(iii)

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

(iv)

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

(v)

\frac{d y}{d x} = 2 x y, y (0) = 1

(vi)

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1

(vii)

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

(viii)

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}

when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)

2 (y + 3) - x y \frac{d y}{d x} = 0

, y(1) = −2 [NCERT EXEMPLAR]
(x)

e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0, y (0) = \frac{π}{4}

[CBSE 2018]

Q. Solve each of the following initial value problems:
(i)

y' + y = e^{x}, y (0) = \frac{1}{2}

(ii)

x \frac{d y}{d x} - y = \log x, y (1) = 0

(iii)

\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0

(iv)

x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0

(v)

(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0

(vi)

\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1

(vii)

x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1

(viii)

\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0

(ix)

\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}

(x)

\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}

(xi)

\frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0

(xii)

d y = \cos x (2 - y \cos e c x) d x

(xiii)

\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y; \tan x \neq 0

given that y = 0 when

x = \frac{π}{2}

Q. Consider the differential equation

. . y + 2 . y + y = 0

with boundary conditions

y (0) = 1

y (1) = 0

. The value of

y (2)

Q. If

y^{'} - 3 y^{'} + 2 y = 0

where

y (0) = 1, y^{'} (0) = 0

, then the value of y at

x = l o g 2

Join BYJU'S Learning Program

It is given that y′′+2y′+y=0,y(0)=0,y(1)=0. What is y(0.5)?

The correct option is A 0Given the differential equation y2+2y′+y=0 (i) Boundary conditions are y(0)=y(1)=0 Auxilary equation of (i) m2+2m+1=0 ⇒m=−1,−1 General soluiton of (i) is y(x)=(A+Bx)e−x Now: y(0)=0 ⇒0=(A+B.0)eo ⇒A=0 and y(1)=0 ⇒0=(0+B.1)e−1 ⇒B=0 y(x)=0 Hence, y(0,5)=0

It is given that $y^{''} + 2 y^{'} + y = 0, y (0) = 0, y (1) = 0$ . What is $y (0.5)$ ?