NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 12 - Differential Equations

At the end of all the chapters of NCERT Class 12 Maths book, there is an exercise known as miscellaneous exercise. This covers questions from the whole chapter. NCERT solutions for Class 12 Maths chapter 9 miscellaneous exercise take a tour through all the concepts covered in this NCERT chapter. Class 12 Maths chapter 9 miscellaneous exercise solutions are a bit more lengthy compared to other exercises of differential equations. Class 12 Maths chapter 9 miscellaneous solutions covers the complete chapter through questions. If one student is able to solve miscellaneous exercise chapter 9 Class 12 without looking to the solution, then it implies he has understood the concepts covered in the chapter. The ideas covered in the last 6 exercises are used in the NCERT solutions for Class 12 Maths chapter 9 miscellaneous exercise. Before solving miscellaneous exercise chapter 9 Class 12 it’s better to cover the following exercises.

  • Differential Equations exercise 9.1

  • Differential Equations exercise 9.2

  • Differential Equations exercise 9.3

  • Differential Equations exercise 9.4

  • Differential Equations exercise 9.5

  • Differential Equations exercise 9.6

Differential Equations Class 12 Chapter 9 -Miscellaneous Exercise

Question:1 Indicate Order and Degree.

(i) \frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x

Answer:

Given function is
\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x
We can rewrite it as
y''+5x(y')^2-6y = \log x
Now, it is clear from the above that, the highest order derivative present in differential equation is y''

Therefore, the order of the given differential equation \frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x is 2
Now, the given differential equation is a polynomial equation in its derivative y '' and y 'and power raised to y '' is 1
Therefore, it's degree is 1

Question:1 Indicate Order and Degree.

(ii) \left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x

Answer:

Given function is
\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x
We can rewrite it as
(y')^3-4(y')^2+7y=\sin x
Now, it is clear from the above that, the highest order derivative present in differential equation is y'

Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives y 'and power raised to y ' is 3
Therefore, it's degree is 3

Question:1 Indicate Order and Degree.

(iii) \frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0

Answer:

Given function is
\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0
We can rewrite it as
y''''-\sin y''' = 0
Now, it is clear from the above that, the highest order derivative present in differential equation is y''''

Therefore, order of given differential equation is 4
Now, the given differential equation is not a polynomial equation in it's dervatives
Therefore, it's degree is not defined

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) xy = ae^x + be^{-x} + x^2\qquad :\ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy +x^2 -2 =0

Answer:

Given,

xy = ae^x + be^{-x} + x^2

Now, differentiating both sides w.r.t. x,

x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x

Again, differentiating both sides w.r.t. x,

\\ (x\frac{d^2y}{dx^2} + \frac{dy}{dx}) + \frac{dy}{dx} = ae^x + be^{-x} + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ae^x + be^{-x} + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy -x^2 + 2 \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 + 2

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(ii) y = e^x(a\cos x + b \sin x )\qquad : \ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

Answer:

Given,

y = e^x(a\cos x + b \sin x )

Now, differentiating both sides w.r.t. x,

\frac{dy}{dx} = e^x(-a\sin x + b \cos x ) + e^x(a\cos x + b \sin x ) =e^x(-a\sin x + b \cos x ) +y

Again, differentiating both sides w.r.t. x,

\\ \frac{d^2y}{dx^2} = e^x(-a\cos x - b \sin x ) + e^x(-a\sin x + b \cos x ) + \frac{dy}{dx} \\ = -y + (\frac{dy}{dx} -y) + \frac{dy}{dx} \\ \implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii) y= x\sin 3x \qquad : \ \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0

Answer:

Given,

y= x\sin 3x

Now, differentiating both sides w.r.t. x,

y= x\sin 3x \frac{dy}{dx} = x(3\cos 3x) + \sin 3x

Again, differentiating both sides w.r.t. x,

\\ \frac{d^2y}{dx^2} = 3x(-3\sin 3x) + 3\cos 3x + 3\cos 3x \\ = -9y + 6\cos 3x \\ \implies \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv) x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0

Answer:

Given,

x^2 = 2y^2\log y

Now, differentiating both sides w.r.t. x,

\\ 2x = (2y^2.\frac{1}{y} + 2(2y)\log y)\frac{dy}{dx} = 2(y + 2y\log y)\frac{dy}{dx} \\ \implies \frac{dy}{dx} = \frac{x}{y(1+ 2\log y)}

Putting \frac{dy}{dx}\ and \ x^2 values in LHS

\\ (2y^2\log y + y^2)\frac{dy}{dx} - xy = y^2(2\log y + 1)\frac{x}{y(1+ 2\log y)} -xy \\ = xy - xy = 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

Question:3 Form the differential equation representing the family of curves given by (x-a)^2 + 2y^2 = a^2 , where a is an arbitrary constant.

Answer:

Given equation is
(x-a)^2 + 2y^2 = a^2
we can rewrite it as
2y^2 = a^2-(x-a)^2 -(i)
Differentiate both the sides w.r.t x
\frac{d\left ( 2y^2 \right )}{dx}=\frac{d(a^2-(x-a)^2)}{dx}
4yy^{'}=4y\frac{dy}{dx}=-2(x-a)\\ \\
(x-a)= -2yy'\Rightarrow a = x+2yy' -(ii)
Put value from equation (ii) in (i)
(-2yy')^2+2y^2= (x+2yy')^2\\ 4y^2(y')^2+2y^2= x^2+4y^2(y')^2+4xyy'\\ y' = \frac{2y^2-x^2}{4xy}
Therefore, the required differential equation is y' = \frac{2y^2-x^2}{4xy}

Question:4 Prove that x^2 - y^2 = c (x^2 + y^2 )^2 is the general solution of differential equation (x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy , where c is a parameter.

Answer:

Given,

(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy

\implies \frac{ dy}{dx} = \frac{(x^3 - 3x y^2 )}{(y^3 - 3x^2 y)}

Now, let y = vx

\implies \frac{ dy}{dx} = \frac{ d(vx)}{dx} = v + x\frac{dv}{dx}

Substituting the values of y and y' in the equation,

v + x\frac{dv}{dx} = \frac{(x^3 - 3x (vx)^2 )}{((vx)^3 - 3x^2 (vx))}

\\\implies v + x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v}\\ \implies x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v} -v = \frac{1 - v^4 }{v^3 - 3v}

\implies (\frac{v^3 - 3v }{1 - v^4})dv = \frac{dx}{x}

Integrating both sides we get,

1517901119530459

Now, 1517901120316962

1517901121031268

Let 151790112181538

151790112256298

\implies 1517901123344265

\implies 1517901124127226

Now, 1517901124894198

1517901125674574

Let v 2 = p

151790112647622

1517901127259961

1517901128042674

1517901128803851

Now, substituting the values of I 1 and I 2 in the above equation, we get,

1517901129585614

Thus,

1517901130366853

1517901131127676

1517901131975586

1517901132764415

\\ (x^2 - y^2)^2 = C'^4(x^2 + y^2 )^4 \\ \implies (x^2 - y^2) = C'^2(x^2 + y^2 )^2 \\ \implies (x^2 - y^2) = K(x^2 + y^2 )^2, where\ K = C'^2

Question:5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

1654780046916 Now, equation of the circle with center at (x,y) and radius r is
(x-a)^2+(y-b)^2 = r^2
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
(x-a)^2+(y-a)^2 = a^2 -(i)
Differentiate it w.r.t x
we will get
2(x-a)+2(y-a)y'= 0\\ \\ 2x-2a+2yy'-2ay' = 0\\ a=\frac{x+yy'}{1+y'} -(ii)
Put value from equation (ii) in equation (i)
(x-\frac{x+yy'}{1+y'})^2+(y-\frac{x+yy'}{1+y'})^2=\left ( \frac{x+yy'}{1+y'} \right )^2\\ \\ (x+xy'-x-yy')^2+(y+yy'-x-yy')^2=(x+yy')^2\\ \\ (y')^2(x-y)^2+(x-y)^2=(x+yy')^2\\ \\ (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2

Question:6 Find the general solution of the differential equation \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0

Answer:

Given equation is
\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0
we can rewrite it as
\frac{dy}{dx } =- \sqrt{\frac{1-y^2}{1-x^2}}\\ \\ \frac{dy}{\sqrt{1-y^2}}= \frac{-dx}{\sqrt{1-x^2}}
Now, integrate on both the sides
\sin^{-1}y + C =- \sin ^{-1}x + C'\\ \\ \sin^{-1}y+\sin^{-1}x= C
Therefore, the general solution of the differential equation \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0 is \sin^{-1}y+\sin^{-1}x= C

Question:7 Show that the general solution of the differential equation \frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0 is given by (x + y + 1) = A (1 - x - y - 2xy) , where A is parameter.

Answer:

Given,

\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0

1517901142938688

1517901143721318

1517901144865249

Integrating both sides,

15179011455756

1517901146296446

1517901147071246

1517901147830720

1517901148634163

1517901149451944

1517901150235484

1517901151028875

1517901151812195

Let 1517901152593257

1517901153375926

Let A = 1517901154163100 ,

1517901154943810

Hence proved.

Question:8 Find the equation of the curve passing through the point \left(0,\frac{\pi}{4} \right ) whose differential equation is \sin x \cos y dx + \cos x \sin y dy = 0.

Answer:

Given equation is
\sin x \cos y dx + \cos x \sin y dy = 0.
we can rewrite it as
\frac{dy}{dx}= -\tan x\cot y\\ \\ \frac{dy}{\cot y}= -\tan xdx\\ \\ \tan y dy =- \tan x dx
Integrate both the sides
\log |\sec y|+C' = -\log|sec x|- C''\\ \\ \log|\sec y | +\log|\sec x| = C\\ \\ \sec y .\sec x = e^{C}
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point \left(0,\frac{\pi}{4} \right )
So,
\sec \frac{\pi}{4} .\sec 0 = e^{C}\\ \\ \sqrt2.1= e^C\\ \\ C = \log \sqrt2
Now,
\sec y.\sec x= e^{\log \sqrt 2}\\ \\ \frac{\sec x}{\cos y} = \sqrt 2\\ \\ \cos y = \frac{\sec x}{\sqrt 2}
Therefore, the equation of the curve passing through the point \left(0,\frac{\pi}{4} \right ) whose differential equation is \sin x \cos y dx + \cos x \sin y dy = 0. is \cos y = \frac{\sec x}{\sqrt 2}

Question:9 Find the particular solution of the differential equation (1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0 , given that y = 1 when x = 0 .

Answer:

Given equation is
(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0
we can rewrite it as
\frac{dy}{dx}= -\frac{(1+y^2)e^x}{(1+e^{2x})}\\ \\ \frac{dy}{1+y^2}= \frac{-e^xdx}{1+e^{2x}}
Now, integrate both the sides
\tan^{-1}y + C' =\int \frac{-e^{x}dx}{1+e^{2x}}
\int \frac{-e^{x}dx}{1+e^{2x}}\\
Put
e^x = t \\ e^xdx = dt
\int \frac{dt}{1+t^2}= \tan^{-1}t + C''
Put t = e^x again
\int \frac{-e^{x}dx}{1+e^{2x}} = -\tan ^{-1}e^x+C''
Put this in our equation
\tan^{-1}y = -\tan ^{-1}e^x+C\\ \tan^{-1}y +\tan ^{-1}e^x=C
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
\\ \tan^{-1}1 +\tan ^{-1}e^0=C\\ \\ \frac{\pi}{4}+\frac{\pi}{4}= C\\ C = \frac{\pi}{2}
Now, put the value of C

\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}
Therefore, the particular solution of the differential equation (1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0 is \tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}

Question:10 Solve the differential equation ye^\frac{x}{y}dx = \left(xe^\frac{x}{y} + y^2 \right )dy\ (y \neq 0)

Answer:

Given,

ye^\frac{x}{y}dx = (xe^\frac{x}{y} + y^2)dy

\\ ye^\frac{x}{y}\frac{dx}{dy} = xe^\frac{x}{y} + y^2 \\ \implies e^\frac{x}{y}[y\frac{dx}{dy} -x] = y^2 \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} = 1

Let \large e^\frac{x}{y} = t

Differentiating it w.r.t. y, we get,

\\ \frac{d}{dy}e^\frac{x}{y} = \frac{dt}{dy} \\ \implies e^\frac{x}{y}.\frac{d}{dy}(\frac{x}{y}) = \frac{dt}{dy} \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} =\frac{dt}{dy}

Thus from these two equations,we get,

\\ \frac{dt}{dy} = 1 \\ \implies \int dt = \int dy \\ \implies t = y + C

1517901178627857

Question:11 Find a particular solution of the differential equation (x - y) (dx + dy) = dx - dy, , given that y = -1 , when x = 0 . (Hint: put x - y = t )

Answer:

Given equation is
(x - y) (dx + dy) = dx - dy,
Now, integrate both the sides
Put
(x-y ) = t\\ dx - dy = dt
Now, given equation become
dx+dy= \frac{dt}{t}
Now, integrate both the sides
x+ y + C '= \log t + C''
Put t = x- y again
x+y = \log (x-y)+ C
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
0+(-1) = \log (0-(-1))+ C\\ C = -1
Now, put the value of C

x+y = \log |x-y|-1\\ \log|x-y|= x+y+1
Therefore, the particular solution of the differential equation (x - y) (dx + dy) = dx - dy, is \log|x-y|= x+y+1

Question:12 Solve the differential equation \left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0) .

Answer:

Given,

\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1

1517901190189935

1517901190951594

This is equation is in the form of 1517901191734385

p = 1517901192495126 and Q = 151790119325976

Now, I.F. = 1517901194040946

We know that the solution of the given differential equation is:

1654780149326

151790119561795

1517901196417819

151790119717945

Question:13 Find a particular solution of the differential equation \frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0) , given that y = 0 \ \textup{when}\ x = \frac{\pi}{2} .

Answer:

Given equation is
\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)
This is \frac{dy}{dx} + py = Q type where p =\cot x and Q = 4xcosec x Q = 4x \ cosec x
Now,
I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C
y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\ \\ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x= \frac{\pi}{2}
at x= \frac{\pi}{2}
0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\ \\ C = - \frac{\pi^2}{2}
Now, put the value of C
y\sin x= 2x^2-\frac{\pi^2}{2}
Therefore, the particular solution is y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)

Question:14 Find a particular solution of the differential equation (x+1)\frac{dy}{dx} = 2e^{-y} -1 , given that y = 0 when x = 0

Answer:

Given equation is
(x+1)\frac{dy}{dx} = 2e^{-y} -1
we can rewrite it as
\frac{e^ydy}{2-e^y}= \frac{dx}{x+1}\\
Integrate both the sides
\int \frac{e^ydy}{2-e^y}= \log |x+1|\\
\int \frac{e^ydy}{2-e^y}
Put
2-e^y = t\\ -e^y dy = dt
\int \frac{-dt}{t}=- \log |t|
put t = 2- e^y again
\int \frac{e^ydy}{2-e^y} =- \log |2-e^y|
Put this in our equation
\log |2-e^y| + C'= \log|1+x| + C''\\ \log (2-e^y)^{-1}= \log (1+x)+\log C\\ \frac{1}{2-e^y}= C(1+x)

Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
\frac{1}{2-e^0}= C(1+0)\\ C = \frac{1}{2}
Now, put the value of C
\frac{1}{2-e^y} = \frac{1}{2}(1+x)\\ \\ \frac{2}{1+x}= 2-e^y\\ \frac{2}{1+x}-2= -e^y\\ -\frac{2x-1}{1+x} = -e^y\\ y = \log \frac{2x-1}{1+x}
Therefore, the particular solution is y = \log \frac{2x-1}{1+x}, x\neq-1

Question:15 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let n be the population of the village at any time t.

According to question,

\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)

\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C

Now, at t=0, n = 20000 (Year 1999)

\\ \implies \log (20000) = k(0) + C \\ \implies C = \log2 + 4

Again, at t=5, n= 25000 (Year 2004)

\\ \implies \log (25000) = k(5) + \log2 + 4 \\ \implies \log 25 + 3 = 5k + \log2 +4 \\ \implies 5k = \log 25 - \log2 -1 =\log \frac{25}{20} \\ \implies k = \frac{1}{5}\log \frac{5}{4}

Using these values, at t =10 (Year 2009)

\\ \implies \log n = k(10)+ C \\ \implies \log n = \frac{1}{5}\log \frac{5}{4}(10) + \log2 + 4 \\ \implies \log n = \log(\frac{25.2.10000}{16}) = \log(31250) \\ \therefore n = 31250

Therefore, the population of the village in 2009 will be 31250.

Question:16 The general solution of the differential equation \frac{ydx - xdy}{y} = 0 is

(A) xy = C

(B) x = Cy^2

(C) y = Cx

(D) y = Cx^2

Answer:

Given equation is
\frac{ydx - xdy}{y} = 0
we can rewrite it as
dx = \frac{x}{y}dy\\ \frac{dy}{y}=\frac{dx}{x}
Integrate both the sides
we will get
\log |y| = \log |x| + C\\ \log \frac{y}{x} = C \\ \frac{y}{x} = e^C\\ \frac{y}{x} = C\\ y = Cx
Therefore, answer is (C)

Question:17 The general solution of a differential equation of the type \frac{dx}{dy} + P_1 x = Q_1 is

(A) ye^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C

(B) ye^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C

(C) xe^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C

(D) xe^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C

Answer:

Given equation is
\frac{dx}{dy} + P_1 x = Q_1
and we know that the general equation of such type of differential equation is

xe^{\int p_1dy} = \int (Q_1e^{\int p_1dy})dy+ C
Therefore, the correct answer is (C)

Question:18 The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is

(A) xe^y + x^2 = C

(B) xe^y + y^2 = C

(C) ye^x + x^2 = C

(D) ye^y + x^2 = C

Answer:

Given equation is
e^x dy + (y e^x + 2x) dx = 0
we can rewrite it as
\frac{dy}{dx}+y=-2xe^{-x}
It is \frac{dy}{dx}+py=Q type of equation where p = 1 \ and \ Q = -2xe^{-x}
Now,
I.F. = e^{\int p dx }= e^{\int 1dx}= e^x
Now, the general solution is
y(I.F.) = \int (Q\times I.F.)dx+C
y(e^x) = \int (-2xe^{-x}\times e^x)dx+C\\ ye^x= \int -2xdx + C\\ ye^x=- x^2 + C\\ ye^x+x^2 = C
Therefore, (C) is the correct answer

More About NCERT Solutions for Class 12 Maths Chapter 9 Miscellaneous Exercise

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Also Read| Differential Equations Class 12th Notes

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