IISER Mathematics - Differential Equations
Exam Duration: 45 Mins Total Questions : 30
The displacement x of a particle moving in a straight line at time t is given by the relation \(x=A\sin { \left( \mu t+k \right) } \) The differential equation representing this phenomenon is
- (a)
\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =-kx\)
- (b)
\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =kx\)
- (c)
\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } ={ u }^{ 2 }x\)
- (d)
\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } -{ u }^{ 2 }x\)
x=A sin(μt+k) --- (i)
\(\frac{dx}{dt}\)=μA cos(μt+k)
\(\frac{d^{2}x}{dt^{2}}=\mu^{2}A sin(\mu t+K)\) --- (ii)
From (i) and (ii), we get
\(\frac{d^{2}x}{dt^{2}}=-\mu ^{2}x\), which is the required differential equation.
The solution of the differential equation \(\left( 1+{ y }^{ 2 } \right) +\left( x-{ e }^{ \tan ^{ -1 }{ y } } \right) \frac { dy }{ dx } =0\), is
- (a)
\(\left( x-2 \right) =k{ e }^{ \tan ^{ -1 }{ y } }\)
- (b)
\(2x{ e }^{ \tan ^{ -1 }{ y } }={ e }^{ \tan ^{ -1 }{ y } }+k\)
- (c)
\(x{ e }^{ \tan ^{ -1 }{ y } }=\tan ^{ -1 }{ y } +k\)
- (d)
\(x{ e }^{ \tan ^{ -1 }{ y } }={ e }^{ \tan ^{ -1 }{ y } }+k\)
The order and degree of the differential equation \(\sqrt { \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } } =\sqrt [ 3 ]{ \frac { dy }{ dx } +5 } \) are respectively
- (a)
2 and 3
- (b)
3 and 2
- (c)
2 and 1
- (d)
2 and 2
Given, \(\sqrt { \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } } =\sqrt [ 3 ]{ \frac { dy }{ dx } +5 } \Longrightarrow { \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 3 }={ \left( \frac { dy }{ dx } +5 \right) }^{ 2 }\)
Order is 2 and degree is 3.
The order and degree of the differential equation \({ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 3/2 }-\sqrt { \left( \frac { { d }y }{ { dx } } \right) } -4=0\) are respectively
- (a)
2 and 6
- (b)
3 and 6
- (c)
1 and 4
- (d)
2 and 4
Given \((\frac{d^{2}y}{dx^{2}})^{\frac{3}{2}}=(\frac{dy}{dx})^{\frac{1}{2}}+4\)
On squaring both sides, we get
\((\frac{d^{2}y}{dx^{2}})^{3}=(\frac{dy}{dx})+8(\frac{dy}{dx})^{\frac{1}{2}}+16\)
\(\Rightarrow \left[ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }-\frac { dx }{ dy } -16 \right] ^{ 2 }=64\)
\(\Rightarrow \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 6 }-32\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }-2\frac { dy }{ dx } \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }+\left( \frac { dy }{ dx } \right) ^{ 2 }-32\frac { dy }{ dx } +256=0\)
So, it is clear that order is 2 and degree is 6.
The solution of differential equation \(\left[ \frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } -\frac { y }{ \sqrt { x } } \right] \frac { dx }{ dy } =1,\left( X\neq 0 \right) \) is
- (a)
\(y{ e }^{ 2\sqrt { x } }=2\sqrt { x } +C\)
- (b)
\(y{ e }^{ 2\sqrt { x } }=3\sqrt { x } +C\)
- (c)
\(2y{ e }^{ 2\sqrt { x } }=3\sqrt { x } +C\)
- (d)
\(y{ e }^{ \sqrt { x } }=2\sqrt { x } +C\)
The given differential equation is
\(\left( \frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } -\frac { y }{ \sqrt { x } } \right) \frac { dx }{ dy } =1\quad \Rightarrow \quad \frac { dy }{ dx } =\frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } -\frac { y }{ \sqrt { x } } \)
\(\Rightarrow \quad \frac { dy }{ dx } +\frac { 1 }{ \sqrt { x } } y=\frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } \)
On comparing with the form \(\frac{dy}{dx}+py=Q\) , we get
\(p=\frac { 1 }{ \sqrt { x } } ,\quad Q=\frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } \)
\(\therefore \quad IF={ e }^{ \int { \frac { 1 }{ \sqrt { x } } dx } }\Rightarrow IF={ e }^{ 2{ x }^{ \frac { 1 }{ 2 } } }={ e }^{ 2\sqrt { x } }\)
The general solution of the given differential equation given by
\(y.IF=\int { Q\times IF\quad dx+C } \)
\(\Rightarrow \quad { ye }^{ 2\sqrt { x } }\times \frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } dx+C\)
\(\Rightarrow \quad { ye }^{ 2\sqrt { x } }=\int { \frac { 1 }{ \sqrt { x } } dx+C } \)
\(\therefore \quad { ye }^{ 2\sqrt { x } }=2\sqrt { x } +C\)
Solve the differential equation \(X\frac { dY }{ dX } =Y\left( \log { Y-\log { X+1 } } \right) \)
- (a)
\(\log { \left( \frac { y }{ x } \right) } ={ x }^{ 2 }C\)
- (b)
\(\log { \left( \frac { y }{ x } \right) } ={ 2x } C\)
- (c)
\(\log { \left( \frac { y }{ x } \right) } ={ x } C\)
- (d)
\(\log { \left( \frac { y }{ x } \right) } ={ y C}\)
Given, \(x\frac { dy }{ dx } =y\quad (logy-logx+1)\)
\(\Rightarrow \quad x\frac { dy }{ dx } =y[log\left( \frac { y }{ x } \right) +1]\)
\(\Rightarrow \quad \frac { dy }{ dx } =\frac { y }{ x } [log\left( \frac { y }{ x } \right) +1]\quad \quad \quad ---\quad (i)\)
This is a homogeneous equation. To simplify it, put y=vx.
\(\Rightarrow \quad \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
On substituting the value of y and \(\frac { dy }{ dx } \) in Eq.(i), we get
\(v+x\frac { dv }{ dx } =\frac { vx }{ x } [log\left( \frac { vx }{ x } \right) +1]\)
\(\Rightarrow \quad x\frac { dv }{ dx } =v[log\quad v+1]-v\)
\(\Rightarrow \quad x\frac { dv }{ dx } =\quad v[log\quad v+1-1]\)
\(\Rightarrow \quad \frac { 1 }{ v\quad logv } dv=\frac { 1 }{ x } dx\)
On integrating both sides, we get
\(\int { \frac { 1 }{ v\quad logv } dv } =\int { \frac { 1 }{ x } dx } \) --- (ii)
Put log v=t \(\frac { 1 }{ v } =\frac { dt }{ dv } \)
⇒ dv= v dt
Then, Eq.(ii) becomes \(\int { \frac { 1 }{ vt } v\quad dt } =\int { \frac { 1 }{ x } dx } \)
\(\Rightarrow \quad \int { \frac { 1 }{ t } dt } =\int { \frac { 1 }{ x } dx } \)
⇒ log(t)=log(x)+log C
⇒log(log v)-logx=log C
⇒ log(log v)-log x=log C [∵ t=log v]
\(\Rightarrow \quad log\left( \frac { log\quad v }{ x } \right) =log\quad C\)
\(\Rightarrow \quad \frac { log\quad v }{ x } =C\Rightarrow \frac { log\left( \frac { y }{ x } \right) }{ x } =C\)
\(\Rightarrow \quad log\left( \frac { y }{ x } \right) =xC\) [∵ y-vx]
Consider the differential equation. \({ Y }^{ 2 }dX+\left( X-\frac { 1 }{ Y } \right) dY=0\) If Y(1) =1, then X is given by
- (a)
\(1-\frac { 1 }{ y } +\frac { { e }^{ 1/y } }{ e } \)
- (b)
\(4-\frac { 2 }{ y } -\frac { { e }^{ 1/y } }{ e } \)
- (c)
\(3-\frac { 1 }{ y } +\frac { { e }^{ 1/y } }{ e } \)
- (d)
\(1+\frac { 1 }{ y } -\frac { { e }^{ 1/y } }{ e } \)
Here, \(\frac{dx}{dy}+\frac{1}{y^{2}}.x=\frac{1}{y^{3}}\) [linear differential equation in x]
Now, \(IF={ e }^{ \int { \frac { 1 }{ { y }^{ 2 } } dy } }={ e }^{ -\frac { 1 }{ y } }\)
∴ Solution is
\(x.{ e }^{ -\frac { 1 }{ y } }=\int { \frac { 1 }{ { y }^{ 3 } } .{ e }^{ -\frac { 1 }{ y } }dy } \Rightarrow x.{ e }^{ -\frac { 1 }{ y } }=\int { \frac { 1 }{ y } .\frac { 1 }{ { y }^{ 2 } } .{ e }^{ -\frac { 1 }{ y } }dy } \)
Put \(-\frac { 1 }{ y } =t\Rightarrow \frac { 1 }{ { y }^{ 2 } } dy=dt\)
\(\therefore \quad { xe }^{ -\frac { 1 }{ y } }=\int { -t.{ e }^{ t }dt } \)
\(\Rightarrow \quad { xe }^{ -\frac { 1 }{ y } }=-[t.{ e }^{ t }-\int { 1.{ e }^{ t }dt } ]+C\)
\(\Rightarrow \quad { xe }^{ -\frac { 1 }{ y } }=-t{ e }^{ t }+{ e }^{ t }+C\)
\(\Rightarrow \quad { xe }^{ -\frac { 1 }{ y } }=\frac { 1 }{ y } .{ e }^{ -\frac { 1 }{ y } }+{ e }^{ -\frac { 1 }{ y } }+C\)
At x=1, y=1, then
e-1=e-1+e-1+C ⇒ C=-\(\frac{1}{e}\)
∴ \({ xe }^{ -\frac { 1 }{ y } }=\frac { 1 }{ y } .{ e }^{ -\frac { 1 }{ y } }+{ e }^{ -\frac { 1 }{ y } }-\frac { 1 }{ e } \Rightarrow x=\frac { 1 }{ y } +1-\frac { 1 }{ e } .{ e }^{ \frac { 1 }{ y } }\)
The differential equation which represents the family of curves, \(y = C_1e^{c_2 x}\) where C1 and C2 are arbitrary constants, is
- (a)
\({ y }^{ \prime }={ y }^{ 2 }\)
- (b)
\({ y }^{ \prime \prime }={ y }^{ \prime }y\)
- (c)
\({ yy }^{ \prime \prime }={ y }^{ \prime }\)
- (d)
\(y{ y }^{ \prime \prime }={ \left( { y }^{ \prime } \right) }^{ 2 }\)
Given, \(y={ C }_{ 1 }{ e }^{ { C }_{ 2 }x }\Rightarrow { y }^{ 1 }={ C }_{ 2 }{ C }_{ 1 }{ e }^{ { C }_{ 2 }x }\)
\(\Rightarrow \quad { y }^{ ' }={ C }_{ 2 }y\) --- (i)
\(\Rightarrow \quad { y }^{ '' }-C_{ 2 }{ y }^{ ' }\)
\(\Rightarrow \quad { y }^{ '' }=\frac { { ({ y }^{ ' }) }^{ 2 } }{ y } \) [from Eq.(i) \({ C }_{ 2 }=\frac { { y }^{ ' } }{ y } \)]
\(\Rightarrow \quad { yy }^{ '' }={ ({ y }^{ ' }) }^{ 2 }\)
The differential equation corresponding to y = \({ c }_{ 1 }e^{ m_{ 1 }x }+{ c }_{ 2 }e^{ m_{ 2 }x }+{ c }_{ 3 }e^{ m_{ 3 }x }\), where c1, c2, c3 are arbitrary constants and m1, m2, m3 are roots of the equation m3 -7m + 6 = 0 is \(a\frac { d^{ 3 }y }{ dx^{ 3 } } +b\frac { d^{ 2 }y }{ dx^{ 2 } } +c\frac { dy }{ dx } +d=0\), where a, b, c, d are constants.
The value of c is
- (a)
6
- (b)
-7
- (c)
2
- (d)
-1
\(\because\) m3 -7m + 6 = 0
\(\Rightarrow\) (m -1)(m - 2)(m + 3) = 0
\(\therefore\) m1 = 1, m2 = 2, m3 = - 3
\(\therefore\) y = c1ex + c2e2x+ c3e3x
\(\Rightarrow\) y1e-x = c1+ c2ex+ c3e-4x
Differentiating both sides w.r.t. x, then
y (-e-x) + e-x \(\frac { dy }{ dx } \) = 0 + c2ex - 4c3e-4x
or -ye-2x + e-2x\(\frac { dy }{ dx } \) = c2 - 4c3e-5x
Again differentiating both sides W.r.t. x, then
(-y ) (-2e-2x) + e-2x \(\left( \frac { -dy }{ dx } \right) \) + e-2x\(\frac { d^{ 2 }y }{ dx^{ 2 } } \)+ \(\frac { dy }{ dx } \)(-2e-2x)
= 0 + 20 c3e-5x
\(\Rightarrow\) 2ye-2x - 3e-2x\(\frac { dy }{ dx } \) + e-2x \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)-= 20C3e-5x
or 2ye3x - 3e3x\(\frac { dy }{ dx } \)+ e3x -\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 20c3
Again differentiating both sides W.r.t. x, then
(2y) (3e3x) + e3x (2\(\frac { dy }{ dx } \)) -3 \(\left\{ e^{ 3x }\frac { d^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } .3e^{ 3x } \right\} \)+ e3x \(\frac { d^{ 3 }y }{ dx^{ 3 } } +\frac { d^{ 2 }y }{ dx^{ 2 } } \).(3e3x)=0
Dividing by e3x, then
6y + 2\(\frac { dy }{ dx } \) - 3\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) - 9\(\frac { dy }{ dx } \) +\(\frac { d^{ 3 }y }{ dx^{ 3 } } \) + 3 \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 0
\(\Rightarrow\)\(\frac { d^{ 3 }y }{ dx^{ 3 } } \) - 7\(\frac { dy }{ dx } \)+ 6y =0
c = -7
The Order of the differential equation is the order of the highest derivative appearing in the equation and the Degree of a differential equation which can be written as polynomial in the derivatives in the degree of the derivative of the highest order occuring in it, after it has been expressed in a form free from radicals and fractions and if differential equation can not be written as a polynomial in the derivatives, then degree deos not defined but order defined.
The order and degree of the differential equation whose solution is y = cx + c3 - 5c3/2 + 3 where c is arbitrary constants, are
- (a)
1 and 3
- (b)
1 and 4
- (c)
1 and 5
- (d)
1 and 6
\(\because \frac { dy }{ dx } =c\)
then, y = x\(\left( \frac { dy }{ dx } \right) +\left( \frac { dy }{ dx } \right) ^{ 3 }-5\left( \frac { dy }{ dx } \right) ^{ 3/2 }+3\)
or \(5\left( \frac { dy }{ dx } \right) ^{ 3/2 }=\left\{ x\frac { dy }{ dx } +\left( \frac { dy }{ dx } \right) ^{ 3 }+3-y \right\} \)
\(\therefore 25\left( \frac { dy }{ dx } \right) ^{ 3 }=\left\{ x\frac { dy }{ dx } +\left( \frac { dy }{ dx } \right) ^{ 3 }+3-y \right\} ^{ 2 }\)
\(\therefore\) Order = 1, degree = 6
Solution of the differential equation \(\left\{ \frac { 1 }{ x } -\frac { { y }^{ 2 } }{ \left( x-y \right) ^{ 2 } } \right\} dx+\left\{ \frac { { x }^{ 2 } }{ \left( x-y \right) ^{ 2 } } -\frac { 1 }{ y } \right\} dy=0\) is
- (a)
\(1n\left| \frac { x }{ y } \right| +\frac { xy }{ (x-y) } =c\)
- (b)
\(1n\left| xy \right| +\frac { xy }{ (x-y) } =c\)
- (c)
\(\frac { xy }{ (x-y) } =c{ e }^{ x/y }\)
- (d)
\(\frac { xy }{ (x-y) } =c{ e }^{ xy }\)
The given equation can be written as
\(\left( \frac { dx }{ x } -\frac { dy }{ y } \right) +\frac { { (x }^{ 2 }dy-{ y }^{ 2 }dx) }{ (x-y)^{ 2 } } =0\)
⇒ \(\left( \frac { dx }{ x } -\frac { dy }{ y } \right) +\frac { \left( \frac { dy }{ { y }^{ 2 } } -\frac { dx }{ { x }^{ 2 } } \right) }{ \left( \frac { 1 }{ y } -\frac { 1 }{ x } \right) ^{ 2 } } =0\)
⇒ \(\left( \frac { dx }{ x } -\frac { dy }{ y } \right) +\frac { \frac { dy }{ { y }^{ 2 } } -\frac { dx }{ { x }^{ 2 } } }{ \left( \frac { 1 }{ x } -\frac { 1 }{ y } \right) ^{ 2 } } =0\)
Integrating, we get
In |x| - In |y| - \(\frac { 1 }{ \left( \frac { 1 }{ x } -\frac { 1 }{ y } \right) } \) = c
⇒ In \(\left| \frac { x }{ y } \right| \) - \(\frac { xy }{ (y-x) } \) = c
or In \(\left| \frac { x }{ y } \right| \) + \(\frac { xy }{ (y-x) } \) = c
If the slope of tangent to the curve is maximum at x = 1 and curve has a minimum value 1 at x = 0, then the curve which also satisfies the equation y'''= 4x - 3 is
- (a)
\(y+2x+\frac { dy }{ dx } =0\)
- (b)
\(y=1+\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 2 } +\frac { { x }^{ 4 } }{ 6 } \)
- (c)
y = 1 + x + x2 + x3
- (d)
none of these
we have, \(\frac { d^{ 3 }y }{ dx^{ 3 } } =4x-3\)
Integrating both sides w.r.t. x, we get
\(\frac { d^{ 2 }y }{ dx^{ 2 } } =2x^{ 2 }-3x+c\)
Given slope of tangent to the curve is maximum at x = 1, then \({ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) }_{ x=1 }\)= 0
∴ 0 = 2 - 3 + c
⇒ c = 1
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 2x2 - 3x + 1
Integrating both sides
\(\frac { dy }{ dx } =\frac { 2x^{ 3 } }{ 3 } -\frac { 3x^{ 2 } }{ 2 } +x+D\)
∵ Curve aas a minimum value 1 at x = 0
∴ \({ \left( \frac { dy }{ dx } \right) }_{ x=0 }\) = 0
⇒ 0 = 0 + D
D = 0
Then, \(\frac { dy }{ dx } =\frac { 2x^{ 3 } }{ 3 } -\frac { 3x^{ 2 } }{ 2 } +x\)
Again integrating both sides, we get
\(y=\frac { { x }^{ 4 } }{ 6 } -\frac { { x }^{ 3 } }{ 2 } +\frac { { x }^{ 2 } }{ 2 } +E\)
Given minimum value of curve is 1 at x=0, then 1 = 0 - 0 + 0 + E
∴ E = 1
Hence, curve is \(y=\frac { { x }^{ 4 } }{ 6 } -\frac { { x }^{ 3 } }{ 2 } +\frac { { x }^{ 2 } }{ 2 } +1\)
Solution of the differential equation (1+y2) dx = (tan-1 y - x )dy is (where c is arbitrary constant)
- (a)
xetan-1 y = (1- tan-1 y) etan -1 y + c
- (b)
xetan-1 y = (tan-1 y - 1) etan -1 y + c
- (c)
x = tan-1 y - 1 + ce-tan-1 y
- (d)
none of the above
We have,
(1 + y2) dx = (tan-1 y - x) dy
⇒ \(\Rightarrow \frac { dx }{ dy } =\frac { tan^{ -1 }y }{ 1+{ y }^{ 2 } } -\frac { x }{ 1+{ { y }^{ 2 } } } \)
\(\frac { dx }{ dy } +\frac { 1 }{ 1+{ { y }^{ 2 } } } .x=\frac { tan^{ -1 }y }{ 1+{ y }^{ 2 } } \)
which is linear differntial equation.
\(\therefore IF={ e }^{ \int { \frac { 1 }{ 1+{ { y }^{ 2 } } } dy } }=e^{ tan-1y }\)
Hence, required solution is
x(IF) = \({ e }^{ \int { \frac { tan^{ -1 }y }{ 1+{ y }^{ 2 } } dy } }\) (IF) dy + c
⇒ \(\Rightarrow { xe }^{ tan-1y }=\int { { e }^{ tan-1y } } .\frac { tan^{ -1 }y }{ (1+y^{ 2 }) } dy+c\)....(i)
Put tan-1 y = t
∴ \(\frac { 1 }{ 1+{ y }^{ 2 } } dy=dt\)
Then, eq (i) reduces to
\(x.{ e }^{ tan-1y }=\int { t.{ e }^{ t }dt+c } \)
= tet- et + c
⇒ xetan -1 y = tan -1y etan-1 y - etan-1 y + c
or xetan -1 y = etan-1 y ( etan-1y -1) + c
(Optioon (b) is correct)
Dividing both sides by e tan-1 y
Then, x = etan-1 y -1 + ce-tan-1 y
(Option (c) is correct)
Let y = (A + Bx) e3x is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } +m\frac { dy }{ dx } +ny=0\), m, n \(\in \) I, then
- (a)
m= -6
- (b)
n = -6
- (c)
m = 9
- (d)
n = 9
∵ y = (A+B x)e3x
or ye-3x = A + Bx
Differentiating both sides w.r.t. x, then we get
y . (-3) e-3x + e-3x \(\frac{dy}{dx}\) = B
Again differentiating both sides w.r.t. x, then
\(\left( -3y+\frac { dy }{ dx } \right) \left( -3e^{ -3x } \right) +{ e }^{ -3x }\left( -3\frac { dy }{ dx } +\frac { d^{ 2 }y }{ dx^{ 2 } } \right) =0\)
Dividing by e-3x, then we get
\(\frac { d^{ 2 }y }{ dx^{ 2 } } -6\frac { dy }{ dx } +9y=0\)
Hence, m=-6, n=9
The degree of the differential equation satisfying \(\sqrt { 1-{ x }^{ 2 } } +\sqrt { 1-{ y }^{ 2 } } =a\left( x-y \right) \), is
- (a)
1
- (b)
2
- (c)
3
- (d)
none of these
Given, a(x - y) = \(\sqrt { 1-{ x }^{ 2 } } +\sqrt { 1-{ y }^{ 2 } } \)
Putting x = sin A, y = sin B, we get
cos A + cos B = a (sin A - sin B)
⇒ 2 \(\cos { \left( \frac { A+B }{ 2 } \right) \cos { \left( \frac { A-B }{ 2 } \right) } } \)
= 2 a \(\cos { \left( \frac { A-B }{ 2 } \right) \sin { \left( \frac { A+B }{ 2 } \right) } } \)
⇒ \(\cot { \left( \frac { A-B }{ 2 } \right) } \) = a ⇒ A - B = 2 cot-1 a.
⇒ sin-1x - sin-1y = 2 cot-1a
On differentiating w.r.t.x, we get
\(\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } -\frac { 1 }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dy }{ dx } =0\quad \Rightarrow \frac { dy }{ dx } =\sqrt { \frac { 1-{ y }^{ 2 } }{ 1-{ x }^{ 2 } } } \)
\({ \left( \frac { dy }{ dx } \right) }^{ 2 }=\frac { 1-{ y }^{ 2 } }{ 1-{ x }^{ 2 } } \Rightarrow \left( 1-{ x }^{ 2 } \right) { \left( \frac { dy }{ dx } \right) }^{ 2 }=\quad 1-{ y }^{ 2 }\)
Clearly, it is differential equation of first order and second degree.
The degree of the differential equation of all curves having normal of constant length c, is
- (a)
1
- (b)
3
- (c)
4
- (d)
none of these
we have, \(y\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } =c\)
⇒ \({ y }^{ 2 }\left\{ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right\} ={ c }^{ 2 }\Rightarrow { y }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ y }^{ 2 }={ c }^{ 2 }\)
Clearly, it is a differential equation of degree 2.
The differential equation having solution as y = 17ex + ae-x is
- (a)
y'' - x = 0
- (b)
y'' - y = 0
- (c)
y' - y = 0
- (d)
y' - x = 0
y' = 17ex - ae-x
y'' = 17ex + ae-x = y ⇒ y'' - y = 0
The differential equation of all non-vertical lines in a plane is
- (a)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\)
- (b)
\(\frac { { d }^{ 2 }x }{ { dy }^{ 2 } } =0\)
- (c)
\(\frac { dy }{ dx } =0\)
- (d)
\(\frac { dx }{ dy } =0\)
The general equation of all non-vertical lines in a plane is ax + by = 1, where b ≠ 0.
⇒ a + \(b\frac { dy }{ dx } =0\) [Differentiating w.r.t. x]
⇒ \(b\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\quad \)[Differentiating w.r.t. x]
⇒\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\) [∵ b ≠ 0]
Hence, the required differential equation is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\).
Given the differential equation \(\frac { dy }{ dx } =\frac { 6{ x }^{ 2 } }{ 2y+\cos { y } } \); y(1) = π. Mark out the correct statement.
- (a)
solution is y2 - sin y = -2x3 + C
- (b)
solution is y2 + sin y = 2x3 + C
- (c)
C = π2 + 2\(\sqrt { 2 } \)
- (d)
C = π2 + 2
We have, \(\int { \left( 2y+\cos { y } \right) dy\quad =\quad \int { 6{ x }^{ 2 }dx } } \)
⇒ y2 + sin y = 2x2 +C
∵ y(1) = π ⇒ C = π2 - 2
∴ Solution is y2 + sin y = 2x3 + π2 - 2
⇒ y2 + sin y = 2x3 + C, where C = π2 - 2
the differential equation \(\frac { dy }{ dx } =\frac { \sqrt { 1-{ y }^{ 2 } } }{ y } \) determines a family of circle with
- (a)
variable radii and fixed centre (0,1)
- (b)
variable radii and fixed centre (0,-1)
- (c)
fixed radius 1 and variable centre on x-axis
- (d)
fixed radius 1 and variable centre on y-axis
We have, \(\frac { ydy }{ \sqrt { 1-{ y }^{ 2 } } } =dx\)
On integration, we get -\(\sqrt { 1-{ y }^{ 2 } } \) = x + c
⇒ 1 - y2 = (x + c)2 ⇒ (x + c)2 + y2 = 1
⇒ radius 1 and centre on the x-axis.
The solution of the differential equation \(\frac { x+y\frac { dy }{ dx } }{ y-x\frac { dy }{ dx } } ={ x }^{ 2 }+2{ y }^{ 2 }+\frac { { y }^{ 4 } }{ { x }^{ 2 } } \) is
- (a)
\(\frac { y }{ 4 } +\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } =c\)
- (b)
\(\frac { 2y }{ x } -\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } =c\)
- (c)
\(\frac { x }{ y } -\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } =c\)
- (d)
none of these
Given, \(\frac { x+y\frac { dy }{ dx } }{ y-x\frac { dy }{ dx } } ={ x }^{ 2 }+2{ y }^{ 2 }+\frac { { y }^{ 4 } }{ { x }^{ 2 } } \quad \)
⇒ \(\frac { d\left( { x }^{ 2 }+{ y }^{ 2 } \right) }{ { \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 } } =2\frac { d\left( \frac { x }{ y } \right) }{ { \left( \frac { x }{ y } \right) }^{ 2 } } \)
Integrating, we get
\(-\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } =\frac { -2 }{ x/y } +c\Rightarrow c=\frac { 2y }{ x } -\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } \)
The solution of the differential equation \(\frac { dy }{ dx } \) = ex-y+ x2e-y is
- (a)
ex = \(\frac { { y }^{ 3 } }{ 3 } \) + ey+ c
- (b)
ey = \(\frac { { x }^{ 2 } }{ 3 } \\ \) + ex + c
- (c)
ey = \(\frac { { x }^{ 3 } }{ 3 } \) + ex + c
- (d)
none of these
From given differential equation
\(\frac { dy }{ dx } =\frac { { e }^{ x }+{ x }^{ 2 } }{ { e }^{ y } } \) ⇒ ey dy = (ex + x2) dx
Integrating, we get ey = ex + \(\frac { { x }^{ 3 } }{ 3 } \)+c
If \(\frac { dy }{ dx } =\frac { x+y }{ x } \) , y(1) = 1, then y =
- (a)
x + ln x
- (b)
x2 + x ln x
- (c)
xex-1
- (d)
x + x ln x
It is homogeneous equation.
Substitute y = vx ⇒ \(\frac { dy }{ dx } =\frac { xdv }{ dx } +v\)
Now, equation becomes
\(\frac { xdv }{ dx } +v=1+v\Rightarrow dv=\frac { dx }{ x } \)
On integration, we get
v = ln x + c ⇒ \(\frac { y }{ x } \) = ln x + c
x = 1, y = 1 ⇒ c = 1 ∴ y = x + x ln x.
The solution of diifferential equation \(\frac { dy }{ dx } =\frac { x-y }{ x+y } \) is
- (a)
x2 - y2 + 2xy + c = 0
- (b)
x2 - y2 - xy + c = 0
- (c)
x2 - y2 + xy + c = 0
- (d)
x2 - y2 - 2xy + c = 0
\(\frac { dy }{ dx } =\frac { x-y }{ x+y } =\frac { 1-y/x }{ 1+y/x } \) .....(i)
Since, ot is a homogeneous differential equation.
Put y = vx ⇒ \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
Hence, eq.(i) becomes
\(v+x\frac { dv }{ dx } =\frac { 1-v }{ 1+v } \Rightarrow x\frac { dx }{ x } =\frac { -1 }{ 2 } \left( \frac { -2-2v }{ 1-2v-{ v }^{ 2 } } \right) dv\)
⇒ \(x\frac { dv }{ dx } =\frac { 1-2v-{ v }^{ 2 } }{ 1+v } \Rightarrow \frac { dx }{ x } =\frac { -1 }{ 2 } \left( \frac { -2-2v }{ 1-2v-{ v }^{ 2 } } \right) dv\)
On intergration, we get
\(\int { \frac { 1 }{ x } } dx=\frac { -1 }{ 2 } \int { \frac { -2-2v }{ 1-2v-{ v }^{ 2 } } } dv\)
⇒ log c1 - 2log x = log (1 - 2v - v2)
⇒ \(\log { \frac { { c }_{ 1 } }{ { x }^{ 2 } } } =\log { \left( \frac { { x }^{ 2 }-2yx-{ y }^{ 2 } }{ { x }^{ 2 } } \right) } \)
⇒ x2 - 2yx - y2 = c1
⇒ x2 - 2yx - y2 + c = 0 (c = -c1)
Which of the following is a second order differntial equation?
- (a)
(y')2 + x = y2
- (b)
y'y'' + y = sin x
- (c)
y''' + (y'')2 + y = 0
- (d)
y' = y2
y'y'' + y = sinx
⇒ \(\left( \frac { dy }{ dx } \right) \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +y=sinx\)
which is second order differential equation.
The degree of the differential equation \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 3 }+6{ y }^{ 5 }=0\) is
- (a)
1
- (b)
2
- (c)
3
- (d)
5
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 3 }+6{ y }^{ 5 }=0\)
Clearly, degree is 1.
Integrating factor of the differential equation \(\frac { dy }{ dx } +y\tan { x } -\sec { x } =0\) is
- (a)
cos x
- (b)
sec x
- (c)
ecosx
- (d)
esecx
\(\frac { dy }{ dx } +y\tan { x } -\sec { x } =0\)
I.F = \({ e }^{ \int { \tan { x } dx } }={ e }^{ \log { \sec { x } } }\) = sec x
the solution of \(x\frac { dy }{ dx } +y={ e }^{ x }\) is
- (a)
\(y=\frac { { e }^{ x } }{ x } +\frac { k }{ x } \)
- (b)
y = xex + cx
- (c)
y = xex + k
- (d)
\(x=\frac { { e }^{ y } }{ y } +\frac { k }{ y } \)
\(x\frac { dy }{ dx } +y={ e }^{ x }\)
\(\frac { dy }{ dx } +\frac { 1 }{ x } y=\frac { { e }^{ x } }{ x } \)
It is a linear differential equation with
I.F. = \({ e }^{ \int { \frac { 1 }{ x } dx } }={ e }^{ \log { x } }\) = x
Now, solution is y.x = \(\int { \frac { { e }^{ x } }{ x } .xdx+k } \)
⇒ yx = ex + k ⇒ y = \(\frac { { e }^{ x } }{ x } +\frac { k }{ x } \).
The order and degree of the differential equation \({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2{ \left( \frac { dy }{ dx } \right) }^{ 4 }={ y }^{ 4 }\) are
- (a)
1, 4
- (b)
3, 4
- (c)
2, 4
- (d)
3, 2
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2{ \left( \frac { dy }{ dx } \right) }^{ 4 }={ y }^{ 4 }\)
Clearly, it is a third order differential equation with degree 2.
Statement-I: 'x' is not an integrating factor for the differential equation \(x\frac { dy }{ dx } +2y={ e }^{ x }\).
Statement-II: \(x\left( x\frac { dy }{ dx } +2y \right) =\frac { d }{ dx } \left( { x }^{ 2 }y \right) \).
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation
of Statement -1. - (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -1.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
\(\frac { dy }{ dx } +\frac { 2 }{ x } y=\frac { { e }^{ x } }{ x } \)
I.F = \({ e }^{ \int { \frac { 2 }{ x } dx } }={ e }^{ \log { { x }^{ 2 } } }\) = x2
⇒ Statement -I is true
Now, \(\frac { d }{ dx } \left( { x }^{ 2 }y \right) ={ x }^{ 2 }\frac { dy }{ dx } +y2x=x\left( x\frac { dy }{ dx } +2y \right) \)
⇒ Statement-II is true.