Differential Equations
Exam Duration: 45 Mins Total Questions : 30
The solution of the first order differential equation \(x^{ \prime }(t)=-3x(t),x(0)={ x }_{ 0 }\) is
- (a)
\(x(t)={ x }_{ 0 }e^{ -3t }\)
- (b)
\(x(t)={ x }_{ 0 }e^{ -3 }\)
- (c)
\({ x }_{ 0 }e^{ -1/3 }\)
- (d)
\({ x }_{ 0 }e^{ -t }\)
\(x(t)={ x }_{ 0 }e^{ -3t }\)
For the equation x"(t)+3x'+2x(t) = 5, the solution x(t) approaches which of the following values as \(t\rightarrow \infty \)?
- (a)
0
- (b)
5/2
- (c)
5
- (d)
10
x = 5/2
The general solution of \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=0\) is
- (a)
y = P cos x + Q sin x
- (b)
y = P cos x
- (c)
y = P sin x
- (d)
y = P \(sin^{ 2 }x\)
Complete solution, y = CF+PI
y = P cos x + Q sin x
The degree of the differential equation \(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +2{ x }^{ 3 }=0\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +2{ x }^{ 3 }=0\)
Degree = The power of highest derivative in differential equation = 1
The solution for the differential equation \(\frac { dy }{ dx } ={ x }^{ 2 }y\) with the condition that y = 1 at x = 0 is
- (a)
\(y={ e }^{ 1/2x }\)
- (b)
\(logy=\frac { { x }^{ 3 } }{ 3 } +4\)
- (c)
\(logy=\frac { { x }^{ 2 } }{ 2 } \)
- (d)
\(y=e^{ x^{ 3 }/3 }\)
\(\frac { dy }{ dx } ={ x }^{ 2 }y\Rightarrow \int { \frac { dy }{ y } } =\int { x } ^{ 2 }dx\\ \Rightarrow logy=\frac { { x }^{ 3 } }{ 3 } +logc\quad (on\quad integration)\\ y=ce^{ x3/3 }\\ Given,\quad y=1,\quad at\quad x=0\\ Then\quad Eq.(i),\quad 1=c{ e }^{ 0 }\\ \Rightarrow \quad c=1\\ y=e^{ x^{ 3 }/3 }\)
The solution of the differential equation \(\frac { dy }{ dx } +\frac { y }{ x } =x\) with the condition that y = 1 at x = 1 is
- (a)
\(y=\frac { 2 }{ { 3x }^{ 2 } } +\frac { x }{ 3 } \)
- (b)
\(y=\frac { x }{ 2 } +\frac { 1 }{ 2x } \)
- (c)
\(y=\frac { 2 }{ 3 } +\frac { x }{ 3 } \)
- (d)
\(y=\frac { 2 }{ 3x } +\frac { { x }^{ 2 } }{ 3 } \)
\(\frac { dy }{ dx } +\frac { y }{ x } =x,y(1)=1\\ IF=e^{ \int { \frac { 1 }{ x } dx } }={ e }^{ logx }=x\\ Complete\quad solution\\ yIF=\int { x } .IFdx\\ yx=\int { { x }^{ 2 }dx+c } \\ xy=\frac { { x }^{ 3 } }{ 3 } +\frac { c }{ x } \Rightarrow y=\frac { { x }^{ 2 } }{ 3 } +\frac { c }{ x } \\ By\quad y(1)=1,\\ c=\frac { 2 }{ 3 } \\ Hence,y=\frac { { x }^{ 2 } }{ 3 } +\frac { 2 }{ 3x } \)
The solution of the ordinary differential equation \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } -6y=0\) is
- (a)
\(y=c_{ 1 }e^{ 3x }+c_{ 2 }e^{ -2x }\)
- (b)
\(y=c_{ 1 }e^{ 3x }+c_{ 2 }e^{ 2x }\)
- (c)
\(c_{ 1 }e^{ -3x }+c_{ 2 }e^{ 2x }\)
- (d)
\(c_{ 1 }e^{ -3x }+c_{ 2 }e^{ -2x }\)
\(\frac { { d }^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } -6y=0\\ \Rightarrow ({ D }^{ 2 }+D-6)y=0,\\ AE\quad is\quad { m }^{ 2 }+m-6=0\\ \Rightarrow { m }^{ 2 }+3m-2m-6=0\\ \Rightarrow \quad (m+3)(m-2)=0\\ \Rightarrow \quad m=-3,2\)
\(c_{ 1 }e^{ -3x }+c_{ 2 }e^{ 2x }\)
The solution of the differential equation \({ x }^{ 2 }\frac { dy }{ dx } +2xy-x+1=0\) given that at x=1, y=0 is
- (a)
\(\frac { 1 }{ 2 } -\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
- (b)
\(\frac { 1 }{ 2 } -\frac { 1 }{ x } -\frac { 1 }{ { 2x }^{ 2 } } \)
- (c)
\(\frac { 1 }{ 2 } +\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
- (d)
\(-\frac { 1 }{ 2 } +\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
\(\frac { 1 }{ 2 } -\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
Biotransformation of an organic compound having concentration x can be modeled using an ordinary differential equation \(\frac { dx }{ dt } +k{ x }^{ 2 }=0\), where k is the reaction rate constant. If x = a at t = 0, the solution of the equation is
- (a)
\(x=a{ e }^{ -kt }\)
- (b)
\(\frac { 1 }{ x } =\frac { 1 }{ a } +kt\)
- (c)
\(x=a(1-e^{ -kt })\)
- (d)
\(x=a+kt\)
\(\frac { 1 }{ x } =\frac { 1 }{ a } +kt\)
The Blasius equation \(\frac { d^{ 3 }f }{ d\eta ^{ 3 } } +\frac { f }{ 2 } \frac { d^{ 2 }f }{ d\eta ^{ 2 } } =0\) is a
- (a)
second order non-linear differential equation
- (b)
third order non-linear ordinary differential equation
- (c)
third order linear ordinary differential equation
- (d)
mixed order non-linear ordinary differential equation
Ordinary Differential Equation (ODE): It is a differential equation in which the unknown function also known as the dependent variable as a function of the single independent variable.
Partial Differential Equation (PDE): It is a differential equation in which the unknown function is a function of multiple independent variables and the equation involves a partial derivative.
Linear and non-linear equation: A differential equation is linear if the known function and its derivatives appear to power 1 and non-linear if power appears to more than 1.
Order of the equation: It is the highest power of derivatives that appears in the differential equation. Hence, the given equation is 3rd order linear ordinary differential equation.
The solution of the differential equation \(\frac { dy }{ dx } +2xy={ e }^{ -x^{ 2 } }\) with y(0) = 1 is
- (a)
\((1+x)e^{ x^{ 2 } }\)
- (b)
\((1+x)e^{ -x^{ 2 } }\)
- (c)
\((1-x)e^{ x^{ 2 } }\)
- (d)
\((1-x)e^{ -x^{ 2 } }\)
\((1+x)e^{ -x^{ 2 } }\)
The solution of the differential equation \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\frac { dy }{ dx } +y=0\) is
- (a)
\(A{ e }^{ x }+Be^{ -x }\)
- (b)
\(e^{ x }(Ax+B)\)
- (c)
\(e^{ x }\left\{ Acos\left( \frac { \sqrt { 3 } }{ 2 } \right) x+Bcos\left( \frac { \sqrt { 3 } }{ 2 } \right) x \right\} \)
- (d)
\(e^{ -x/2 }\left\{ Acos\left( \frac { \sqrt { 3 } }{ 2 } \right) x+Bsin\left( \frac { \sqrt { 3 } }{ 2 } \right) x \right\} \)
Given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\frac { dy }{ dx } +y=0\)
AE is \({ m }^{ 2 }+m+1=0\)
\(m=-\frac { 1 }{ 2 } \pm \frac { \sqrt { 3 } }{ 2 } i\\ y=CF+PI\)
\(e^{ -x/2 }\left\{ Acos\left( \frac { \sqrt { 3 } }{ 2 } \right) x+Bsin\left( \frac { \sqrt { 3 } }{ 2 } \right) x \right\} \)
For \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } +4\frac { dy }{ dx } +3y=3e^{ 2x }\) , the particular integral is
- (a)
\(\frac { 1 }{ 15 } e^{ 2x }\)
- (b)
\(\frac { 1 }{ 5 } e^{ 2x }\)
- (c)
\(3e^{ 2x }\)
- (d)
\(c_{ 1 }e^{ -x }+c_{ 1 }e^{ -3x }\)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\frac { dy }{ dx } +y=0\\ AE\quad is\quad { m }^{ 2 }+m+1=0\\ m=-\frac { 1 }{ 2 } \pm \frac { \sqrt { 3 } }{ 2 } i\\ CF={ e }^{ x/2 }\left\{ Acos\frac { \sqrt { 3 } }{ 2 } x+Bsin\frac { \sqrt { 3 } }{ 2 } x\quad and\quad PI=0 \right\} \\ y=CF+PI\)
\(\frac { 1 }{ 5 } e^{ 2x }\)
The complete solution for the ordinary differential equation
\(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +qy=0\quad is\quad y={ c }_{ 1 }e^{ -x }+{ c }_{ 2 }e^{ -3x }\)
Then, p and q are
- (a)
p = 3 and q = 3
- (b)
p = 3 and q = 4
- (c)
p = 4 and q = 3
- (d)
p = 4 and q = 4
Given, \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +qy=0\)
Its solution is \(y={ c }_{ 1 }e^{ -x }+{ c }_{ 2 }e^{ -3x }\)
Given equation can be written as
\(({ D }^{ 2 }+pD+q)y=0\\ { D }^{ 2 }+pD+q=0\)
From the solution of equation, we get roots of equation are -1 and -3.
Sum of roots
-p = -1+(-3)
p = 4
Product of roots = q = -1 * (-3) = 3
The complete solution for the ordinary differential equation
\(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +qy=0\quad is\quad y={ c }_{ 1 }e^{ -x }+{ c }_{ 2 }e^{ -3x }\)
Which of the following is a solution of the differential equation \(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +(q+1)y=0\) ?
- (a)
\(e^{ -3x }\)
- (b)
\(xe^{ -x }\)
- (c)
\(xe^{ -2x }\)
- (d)
\(x^{ 2 }e^{ -2x }\)
Given equation is \(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +(q+1)y=0\)
As p = 4, q = 3 from previous question
\(\frac { { d }^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +qy=0\\ y={ c }_{ 1 }e^{ -x }+{ c }_{ 2 }e^{ -3x }\\ ({ D }^{ 2 }+4D+4)y=0\\ AE\quad is\quad { m }^{ 2 }+4m+4=0\\ (m+2)^{ 2 }=0\\ m=-2,-2\)
\(xe^{ -2x }\)
The solution of the differential equation \(\frac { dy }{ dx } +{ y }^{ 2 }=0\) is
- (a)
\(y=\frac { 1 }{ x+c } \)
- (b)
\(y=-\frac { { x }^{ 3 } }{ 3 } +c\)
- (c)
\(ce^{ x }\)
- (d)
Unsolvable as equation is non-linear
Given, \(\frac { dy }{ dx } +{ y }^{ 2 }=0\)
\(\Rightarrow \frac { dy }{ dx } =-{ y }^{ 2 }\\ \Rightarrow \frac { dy }{ { y }^{ 2 } } =-dx\)
Integrate the equation on both sides, we get
\(\int { \frac { dy }{ { y }^{ 2 } } } =-\int { dx } \\ \Rightarrow -\frac { 1 }{ 2 } =-x-c\\ \Rightarrow \frac { 1 }{ y } =x+c\\ \Rightarrow y=\frac { 1 }{ x+c } \)
The solution of the differential equation \(\frac { dy }{ dx } =ky,\) y(0) = c is
- (a)
\(x=ce^{ -ky }\)
- (b)
\(x=ke^{ cy }\)
- (c)
\(y=ce^{ ky }\)
- (d)
\(y=ce^{ -kx }\)
Given, y(0) = c and \(\frac { dy }{ dx } =ky,\)
\(\Rightarrow \frac { dy }{ y } =kdx\\ logy=kx+{ c }_{ 1 }\\ \Rightarrow y=e^{ kx }e^{ c1 }\\ when\quad y(0)=c\)
\(y=ce^{ ky }\)
A function n(x) satisfied the differential equation \(\frac { { d }^{ 2 }n(x) }{ dx^{ 2 } } -\frac { n(x) }{ { L }^{ 2 } } =0\), where L is a constant. The boundary conditions are n(0) = k and \(n(\infty )=0\). The solution to this equation is
- (a)
\(n(x)=ke^{ x/L }\)
- (b)
\(n(x)=ke^{ -x/\sqrt { L } }\)
- (c)
\(n(x)=k^{ 2 }e^{ -x/L }\)
- (d)
\(n(x)=ke^{ -x/L }\)
Given differential equation is
\(\frac { { d }^{ 2 }n(x) }{ d{ x }^{ 2 } } -\frac { n(x) }{ { L }^{ 2 } } =0\\ \Rightarrow \left( D^{ 2 }-\frac { 1 }{ { L }^{ 2 } } \right) n(x)=0\\ AE\quad is\quad { m }^{ 2 }-\frac { 1 }{ { L }^{ 2 } } =0\\ m=-\frac { 1 }{ L } ,+\frac { 1 }{ L } \\ \Rightarrow n(x)=A{ e }^{ -x/L }+B{ e }^{ x/L }\\ \because \quad n(\infty )=0\\ B=0\\ and\quad n(0)=k\\ A=k\\ \therefore \quad n(x)=ke^{ -x/L }\)
Which of the following is a solution to the differential equation \(\frac { dx(t) }{ dt } +3x(t)=0?\)
- (a)
\(x(t)=3{ e }^{ -t }\)
- (b)
\(x(t)=2{ e }^{ -3t }\)
- (c)
\(x(t)=-\frac { 3 }{ 2 } { t }^{ 2 }\)
- (d)
\(x(t)=3{ t }^{ 2 }\)
\(x(t)=2{ e }^{ -3t }\)
The following differential equation ha \(3\left( \frac { { d }^{ 2 }y }{ d{ t }^{ 2 } } \right) +4\left( \frac { dy }{ dt } \right) ^{ 3 }+{ y }^{ 2 }+2=x\)
- (a)
degree = 2 and order = 1
- (b)
degree = 1 and order = 2
- (c)
degree = 4 and order = 3
- (d)
degree = 2 and order = 3
From the definition of order and degree of differential equation order = 2 and degree = 1.
A solution of the following differential equation is given by \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ dx } +6y=0\)
- (a)
\(y={ e }^{ 2x }+{ e }^{ -3x }\)
- (b)
\(y={ e }^{ 2x }+{ e }^{ 3x }\)
- (c)
\(y={ e }^{ -2x }+{ e }^{ 3x }\)
- (d)
\(y={ e }^{ -2x }+{ e }^{ -3x }\)
\(({ D }^{ 2 }-5D+6)y=0\\ \therefore \quad AE\quad is\quad { D }^{ 2 }-5D+6=0\\ { m }^{ 2 }=5m+6=0\\ (d-2)(d-3)=0\\ d=2,3\\ y=e^{ 2x }+e^{ 3x }\)
The solution of differential equation \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\left( 9x-\frac { 20 }{ { x }^{ 2 } } \right) y=0\) in terms of Bessel's function is
- (a)
\(y=\sqrt { x } \left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })+{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
- (b)
\(y=x\left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })+{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
- (c)
\(y=\sqrt { x } \left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })-{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
- (d)
\(y=x^{ 2 }\left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })-{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
\(y=\sqrt { x } \left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })+{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
A tightly stretched string with fixed end points x = 0 and x = pi is initially at rest in its equilibrium position. It is set viberating by giving t each of its points an initial velocity \(\left( \frac { \partial y }{ \partial t } \right) _{ t=0 }=0.03sinx-0.04sin3x\) What will be then find the displacement y(x, t) at any point of string at any time t?
- (a)
\(\frac { 1 }{ c } \) [0.03 sin x sin ct - 0.0133 sin 3x sin 3ct]
- (b)
\(\frac { 1 }{ c } \)[0.03 sin 3x sin 3ct - 0.0133 sin x sin ct]
- (c)
\(\frac { 1 }{ c } \) [0.05 sin x sin ct - 0.0133 sin 3x sin 3ct]
- (d)
\(\frac { 1 }{ c } \) [0.0133 sin x sin ct - 0.03 sin 3x sin 3ct]
\(\frac { 1 }{ c } \) [0.03 sin x sin ct - 0.0133 sin 3x sin 3ct]
The solution of \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +2\frac { dy }{ dx } +17y=0\); y(0) = 1, \(\frac { dy }{ dx } \left( \frac { \pi }{ 4 } \right) =0\) in the range 0 < x < \(\frac { \pi }{ 4 } \) is given by
- (a)
\(e^{ -x }\left( \cos { 4x } +\frac { 1 }{ 4 } \sin { 4x } \right) \)
- (b)
\(e^{ -x }\left( \cos { 4x } -\frac { 1 }{ 4 } \sin { 4x } \right) \)
- (c)
\(e^{ -4x }\left( \cos { 4x } +\frac { 1 }{ 4 } \sin { 4x } \right) \)
- (d)
\(e^{ -4x }\left( \cos { 4x } -\frac { 1 }{ 4 } \sin { 4x } \right) \)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +2\frac { dy }{ dx } +17y=0\\ y(0)=1,\frac { dy }{ dx } \left( \frac { \pi }{ 4 } \right) =0\quad \quad \left( 0<x<\frac { \pi }{ 4 } \right) \)
\(\Rightarrow \quad \left( { D }^{ 2 }+2D+17 \right) y=0\)
AE is \({ m }^{ 2 }+2m+17=0\)
\(m=\frac { -2\pm 8i }{ 2 } =-1\pm 4i\)
\(CF={ e }^{ -x }\left( { c }_{ 1 }\cos { 4x } +{ c }_{ 2 }\sin { 4x } \right) \\ PI=0\)
y = CF+ PI
At x = 0 and y = 1,
1 = \({ e }^{ 0 }\left( { c }_{ 1 }\cos { 0 } +{ c }_{ 2 }\sin { 0 } \right) \)
c1 = 1
From Eq. (i),
\(\frac { dy }{ dx } =-{ e }^{ -x }\left( { c }_{ 1 }\cos { 4x } +{ c }_{ 2 }\sin { 4x } \right) +{ e }^{ -x }\left( { -4c }\sin { 4x } +4{ c }_{ 2 }\cos { 4x } \right) \)
At \(\frac { dy }{ dx } =0\quad and\quad x=\frac { \pi }{ 4 } \),
\(0=-{ e }^{ -\pi /4 }\left( -{ c }_{ 1 }+0 \right) +{ e }^{ -\pi /4 }\left( 0-4{ c }_{ 2 } \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \because \cos { \pi } =-1,\sin { \pi } =0 \right] \)
c1-4c2=0 \(\Rightarrow \quad { c }_{ 2 }=\frac { 1 }{ 4 } \)
Hence, the solution is
y = \(e^{ -x }\left( \cos { 4x } +\frac { 1 }{ 4 } \sin { 4x } \right) \)
The solution of \(x\frac { dy }{ dx } +y={ x }^{ 4 }\) with the condition \(y(1)=\frac { 6 }{ 5 } \) is
- (a)
\(y=\frac { { x }^{ 4 } }{ 5 } +\frac { 1 }{ x } \)
- (b)
\(y=\frac { { 4x }^{ 4 } }{ 5 } +\frac { 4 }{ 5x } \)
- (c)
\(y=\frac { { x }^{ 4 } }{ 5 } +1\)
- (d)
\(y=\frac { { x }^{ 5 } }{ 5 } +1\)
The given differential equation
\(x\frac { dy }{ dx } +y={ x }^{ 4 }\)
\(\frac { dy }{ dx } +\frac { 1 }{ x } .y={ x }^{ 3 }\)
IF = \({ e }^{ \int { \frac { 1 }{ x } dx } }\)
\(\Rightarrow \quad \quad { e }^{ \log { x } }=x\)
and complete solution is
\(y(IF)=\int { { x }^{ 3 } } \left( IF \right) dx+c\)
\(yx=\int { { x }^{ 3 } } x\quad dx+c\)
\(\Rightarrow \quad \quad \int { { x }^{ 4 } } dx+c\)
\(y.x=\frac { { x }^{ 5 } }{ 5 } +c\)
At x = 1 and \(y=\frac { 6 }{ 5 } \)
From Eq.(i),
\(xy=\frac { { x }^{ 5 } }{ 5 } =\frac { 5 }{ 5 } \\ y=\frac { { x }^{ 4 } }{ 5 } +\frac { 1 }{ 4 } \)
It is given that y'' + 2y' + y = 0, y (0) = 0, y (1) = 0. What is y (0.5)?
- (a)
0
- (b)
0.37
- (c)
0.62
- (d)
1.13
Given, y'' + 2y' + y = 0
y (0) = 0
y (1) = 0
y (0.5) = ?
AE is m2 + 2m + 1 = 0
(m+ 1) = 0
m = -1, -1
y= (c1 + c2x)e-x .......(i)
y (0) = 0
0 = c1 + 0
c1 = 0
y (1) = 0
0 = 0 + c2e-1
c2 = 0 \(\left( \because { c }_{ 1 }=0 \right) \)
\(\Rightarrow \) y = e-x0 = 0
y (0.5) = 0
If \({ x }^{ 2 }\frac { dy }{ dx } +2xy=\frac { 2\log { x } }{ x } \)and y(1) = 0, then what
- (a)
e
- (b)
1
- (c)
\(\frac { 1 }{ e } \)
- (d)
\(\frac { 1 }{ { e }^{ 2 } } \)
Given, \({ x }^{ 2 }\frac { dy }{ dx } +2xy=\frac { 2\log { x } }{ x } \)
\(\frac { dy }{ dx } +\frac { 2y }{ x } =\frac { 2\log { x } }{ { x }^{ 3 } } \Rightarrow \frac { dy }{ dx } +Py=Q\)
\(\therefore \quad \quad P=\frac { 2 }{ x } \quad and\quad Q=\frac { 2\log { x } }{ { x }^{ 3 } } \)
\(IF={ e }^{ \int { P } dx }={ e }^{ \int { \frac { 2 }{ x } } dx }={ x }^{ 2 }\)
Complete solution is y(IF)=\(\int { Q(IF) } dx+c\)
\(\Rightarrow \quad y{ x }^{ 2 }=\int { \frac { 2\log { x } }{ { x }^{ 3 } } } { x }^{ 2 }.dx+c\\ \Rightarrow \quad y{ .x }^{ 2 }={ \left( \log { x } \right) }^{ 2 }+c\)
Given, x = 1, y = 0
Then, c = 0
\({ x }^{ 2 }y={ \left( \log { x } \right) }^{ 2 }\)
y=\(\frac { 1 }{ { e }^{ 2 } } \)
The solution of the differential equation \(k^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y-y_{ 2 }\) under boundary conditions.
\((i)\quad y=y_{ 1 }\quad at\quad x=0\\ (ii)\quad y=y_{ 2 }\quad at\quad x=\infty ,\quad where\quad { k }_{ 1 }{ y }_{ 1 }\quad and\quad y_{ 2 }\quad are\quad constants\)
- (a)
\(y=({ y }_{ 1 }-{ y }_{ 2 })e^{ (-x/k^{ 2 }) }+y_{ 2 }\)
- (b)
\(y={ y }_{ 1 }-{ y }_{ 2 }\quad e^{ (-x/k^{ 2 }) }+y_{ 1 }\)
- (c)
\(y=({ y }_{ 1 }-{ y }_{ 2 })sinh\left( \frac { x }{ k } \right) +y_{ 1 }\)
- (d)
\(y=({ y }_{ 1 }-{ y }_{ 2 })e^{ (-x/k }+y_{ 2 }\)
The given differential equation is
\(k^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y-y_{ 2 }\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { y }{ k^{ 2 } } =-\frac { y_{ 2 } }{ k^{ 2 } } \\ AE\quad is\quad m^{ 2 }-\frac { 1 }{ k^{ 2 } } =0\Rightarrow m=\pm \frac { 1 }{ k } \\ CF\quad y={ c }_{ 1 }e^{ x/k }+c_{ 2 }e^{ -x/k }\\ PI=\frac { 1 }{ D^{ 2 }-\frac { 1 }{ k^{ 2 } } } \left( -\frac { y_{ 2 } }{ k^{ 2 } } \right) =-\frac { y_{ 2 } }{ k^{ 2 } } .\frac { 1 }{ D^{ 2 }-\frac { 1 }{ k^{ 2 } } } e^{ 0x }\\ =-\frac { y_{ 2 } }{ k^{ 2 } } .(-k^{ 2 })=y_{ 2 }\\ Complete\quad solution\quad is\quad y=c_{ 1 }e^{ x/k }+c_{ 2 }e^{ -x/k }+y_{ 2 }\\ At\quad x=0,\quad y=y_{ 1 },\quad y_{ 1 }=c_{ 1 }+{ c }_{ 2 }+y_{ 2 }\\ At\quad x=\infty ,\quad y=y_{ 2 },\\ \Rightarrow \quad \quad { c }_{ 1 }=0\Rightarrow { c }_{ 2 }={ y }_{ 1 }-{ y }_{ 2 }\\ y=({ y }_{ 1 }-{ y }_{ 2 }){ e }^{ -x/k }+{ y }_{ 2 }\)
For the differential equation \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +k^{ 2 }y=0\) the boundary conditions are
\((i)\quad y=0,\quad for\quad x=0\\ (ii)\quad y=0,\quad for\quad x=a\)
- (a)
\(y=\underset { m }{ \Sigma } A_{ m }sin\frac { m\pi x }{ a } \)
- (b)
\(y=\underset { m }{ \Sigma } A_{ m }cos\frac { m\pi x }{ a } \)
- (c)
\(y=\underset { m }{ \Sigma } A_{ m }x^{ m\pi /a }\)
- (d)
\(y=\underset { m }{ \Sigma } A_{ m }e^{ -\frac { m\pi x }{ a } }\)
\(AE\quad is\quad m^{ 2 }+k^{ 2 }=0\\ m=\pm ik\\ CF\quad y=c_{ 1 }\quad cos\quad kx+c_{ 2 }sinkx\\ At\quad x=0,\quad y=0\quad \Rightarrow c_{ 1 }=0\\ At\quad x=a,\quad y=0,\\ 0=0+{ c }_{ 2 }sinka\\ \Rightarrow sinks=0\\ (\because c_{ 2 }\neq 0,\quad otherwise\quad y=0\quad will\quad hold\quad always)\\ \Rightarrow \quad \quad ka=m\pi \\ k=\frac { m\pi }{ a } \\ \therefore y=\underset { m }{ \Sigma } A_{ m }sin\frac { m\pi x }{ a } \)
In the recurrence relation the value of in nPn (x) is equal to
- (a)
(2n+1)xPn+1(x)-(n-1)Pn-2(x)
- (b)
(2n-1)xPn-1(x)-(n-1)Pn-2(x)
- (c)
(2n-1)xPn-1(x)+(n-1)Pn-2(x)
- (d)
xP'n(x)-P'n-1(x)
\(We\quad know\quad that\\ (1-2xh+h^{ 2 })^{ -1/2 }=\overset { \infty }{ \underset { h=0 }{ \Sigma } } h^{ n }P_{ n }(x)\quad ....(i)\\ Differential\quad both\quad sides\quad w.r.t.\quad h,\\ -\frac { 1 }{ 2 } (1-2xh+h^{ 2 })^{ -3/2 }(2h-2x)=\overset { \infty }{ \underset { h=0 }{ \Sigma } } nh^{ n-1 }P_{ n }(x)\\ \Rightarrow (x-h)(1-2xh+h^{ 2 })^{ -1/2 }=(1-2xh+h^{ 2 })\\ \overset { \infty }{ \underset { 0 }{ \Sigma } nh } ^{ n-1 }P_{ n(x) }=(1-2xh+h^{ 2 })\overset { \infty }{ \underset { 0 }{ \Sigma } nh } ^{ n-1 }P_{ n }(x)\\ Equating\quad coefficient\quad of\quad h^{ n-1 }\quad on\quad boh\quad sides,\\ xP_{ n-1 }(x)-P_{ n-2 }(x)=nP_{ n }(x)-2x(n-1)P_{ n-1 }(x\_ +(n-2)P_{ n-2 }(x)\\ nP_{ n }(x)=(2n-1)xP_{ n-1 }(x)-(n-1)P_{ n-2 }(x)\)