A function nx satisfies the differential equaation d2nxdx2-nxL2-0 where L is a constant- The boundary conditions are- n0-K and n-0- The soluiton to this equation is

D^2n(x)dx^2 n(x)L^2=0 A.E is D^2 1L^2=0 D=±1L n=c_1e^x/L+c_2e^ x/L n(0)=K ⇒ c_1+c_2=K ..... (1) n(∞)=0 i.e.,0=c_1e^∞+0 ⇒ c_1=0 (1)⇒ c_2=k n(x)=ke^ x/L ...

You visited us 5 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Engineering Mathematics

Homogeneous Linear Differential Equations (General Form of LDE)

A function nx...

Question

A function $n (x)$ satisfies the differential equaation $\frac{d^{2} n (x)}{d x^{2}} - \frac{n (x)}{L^{2}} = 0$ where $L$ is a constant. The boundary conditions are: $n (0) = K$ and $n (\infty) = 0$ . The soluiton to this equation is

n (x) = K exp (x / L)

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

n (x) = K exp (- x / \sqrt{L})

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

n (x) = K^{2} exp (- x / L)

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

n (x) = K exp (- x / L)

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

Open in App

Solution

The correct option is D $n (x) = K exp (- x / L)$
$\frac{d^{2} n (x)}{d x^{2}} - \frac{n (x)}{L^{2}} = 0$
A.E is $D^{2} - \frac{1}{L^{2}} = 0$
$D = \pm \frac{1}{L}$
$n = c_{1} e^{x / L} + c_{2} e^{- x / L}$
$n (0) = K$
$\Rightarrow c_{1} + c_{2} = K$ ..... (1)
$n (\infty) = 0$
i.e., $0 = c_{1} e^{\infty} + 0$
$\Rightarrow c_{1} = 0$
(1) $\Rightarrow c_{2} = k$
$n (x) = k e^{- x / L}$

Suggest Corrections

Similar questions

Q. A function

n (x)

satisfies the differential equaation

\frac{d^{2} n (x)}{d x^{2}} - \frac{n (x)}{L^{2}} = 0

where

L

is a constant. The boundary conditions are:

n (0) = K

and

n (\infty) = 0

. The soluiton to this equation is

Q. The solution to the differential equation

\frac{d^{2} u}{d x^{2}} - k \frac{d u}{d x} = 0

where k is constant, subjected to the boundary conditions

u (0) = 0

and

u (L) = U

, is

Q. Show that

y = A cos n x + B sin n x

, is a solution of the differential equation:

\frac{d^{2} y}{d x^{2}} + n^{2} y = 0

Q. The soluitons of the differential equations

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

are

Q. Consider the differential equaion

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - 4 y = 0

with the boundary condition of

y (0) = 0

and

y (1) = 1

. The complete solution of the differential equation is

Join BYJU'S Learning Program

A function n(x) satisfies the differential equaation d2n(x)dx2−n(x)L2=0 where L is a constant. The boundary conditions are: n(0)=K and n(∞)=0. The soluiton to this equation is

The correct option is D n(x)=Kexp(−x/L)d2n(x)dx2−n(x)L2=0 A.E is D2−1L2=0 D=±1L n=c1ex/L+c2e−x/L n(0)=K ⇒c1+c2=K ..... (1) n(∞)=0 i.e.,0=c1e∞+0 ⇒c1=0 (1)⇒c2=k n(x)=ke−x/L

A function $n (x)$ satisfies the differential equaation $\frac{d^{2} n (x)}{d x^{2}} - \frac{n (x)}{L^{2}} = 0$ where $L$ is a constant. The boundary conditions are: $n (0) = K$ and $n (\infty) = 0$ . The soluiton to this equation is