Class 12 NCERT Maths – Exercise 9.3 Solutions, Chapter 9: Differential Equations (Variable Seperable)

Q1: Find the general solution of the differential equation: \frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}

Solution:
Step 1: Simplify the Trigonometric Expression
Recall the identities:

1 - \cos x = 2 \sin^2(\frac{x}{2})
1 + \cos x = 2 \cos^2(\frac{x}{2})

Substitute these into the differential equation:
\frac{dy}{dx} = \frac{2 \sin^2(\frac{x}{2})}{2 \cos^2(\frac{x}{2})}

\frac{dy}{dx} = \tan^2(\frac{x}{2}) … (1)

Step 2: Separate the Variables
Move the dx term to the right side to separate the variables :
dy = \tan^2(\frac{x}{2}) dx

Step 3: Integrate Both Sides
\int dy = \int \tan^2(\frac{x}{2}) dx

\int dy = \int [\sec^2(\frac{x}{2}) - 1] dx

Step 4: Perform the Integration

• Integral of 1 with respect to y is y .
• Integral of \sec^2(\frac{x}{2}) is \frac{\tan(x/2)}{1/2} = 2 \tan(\frac{x}{2}) .
• Integral of -1 with respect to x is -x .

So, the equation becomes:
y = 2 \tan(\frac{x}{2}) - x + C

Final Answer: The general solution is \boxed{ y = 2 \tan(\frac{x}{2}) - x + C } .

Here are the step-by-step solutions for Questions 2 to 6 of Exercise 9.3, formatted exactly for your WordPress post.

Q2: Find the general solution of the differential equation: \frac{dy}{dx} = \sqrt{4-y^2} \quad (-2 < y < 2)

Solution:
Step 1: Separate the Variables
The given differential equation is: \frac{dy}{dx} = \sqrt{4-y^2}

\frac{dy}{\sqrt{4-y^2}} = dx

Step 2: Integrate Both Sides
\int \frac{dy}{\sqrt{4-y^2}} = \int dx

Recall the standard integral formula: \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a}) .
Here, a = 2 .

Step 3: Perform the Integration
\sin^{-1}(\frac{y}{2}) = x + C

Step 4: Express explicitly in terms of y
\frac{y}{2} = \sin(x + C)

y = 2 \sin(x + C)

Final Answer: The general solution is \boxed{ y = 2 \sin(x + C) } .

Q3: Find the general solution of the differential equation: \frac{dy}{dx} + y = 1 \quad (y \ne 1)

Solution:
Step 1: Separate the Variables:
\frac{dy}{dx} = 1 - y

\frac{dy}{1-y} = dx

Step 2: Integrate Both Sides
\int \frac{dy}{1-y} = \int dx

Step 3: Perform the Integration
Recall that \int \frac{1}{x} dx = \log|x| . Be careful with the coefficient of y , which is -1 .

-\log|1-y| = x + C_1

Multiply by -1 :
\log|1-y| = -1(x + C_1)

\log|1-y| = -x - C_1

Step 4: Simplify the Solution Convert from logarithmic form to exponential form:
|1-y| = e^{-x - C_1}

1-y = \pm e^{-C_1} e^{-x}

1-y = A e^{-x}

y = 1 - A e^{-x}
(Note: Since A is an arbitrary constant, -A is also just a constant. This can also be written as y = 1 + A e^{-x} ).

Final Answer: The general solution is \boxed{ y = 1 + A e^{-x} } .

Q4: Find the general solution of the differential equation: \sec^2 x \tan y , dx + \sec^2 y \tan x , dy = 0

Solution:
Step 1: Separate the Variables
We need to group x-terms with dx and y-terms with dy.
\sec^2 x \tan y dx = - \sec^2 y \tan x dy

\frac{\sec^2 x}{\tan x} dx = - \frac{\sec^2 y}{\tan y} dy

Step 2: Integrate Both Sides
\int \frac{\sec^2 x}{\tan x} dx = - \int \frac{\sec^2 y}{\tan y} dy

Step 3: Perform the Integration
Notice that the numerator is the derivative of the denominator ( \frac{d}{dx}(\tan x) = \sec^2 x ).

Using substitution u = \tan x , \implies du = \sec^2 x dx
Using substitution v = \tan y , \implies dv = \sec^2 y dy

\int \frac{du}{u} dx = - \int \frac{dv}{v} dy

\log {u} dx = - \log {v} dy

\log|\tan x| + \log|\tan y| = \log C
(Note: We use \log C instead of just C to simplify the logarithmic terms easily).

Step 4: Simplify Using Log Properties
Using \log m + \log n = \log mn :

\log |\tan x \tan y| = \log C

\tan x \tan y = C

Final Answer: The general solution is \boxed{ \tan x \tan y = C } .

Q5: Find the general solution of the differential equation: (e^x + e^{-x}) dy - (e^x - e^{-x}) dx = 0

Solution:
Step 1: Separate the Variables
(e^x + e^{-x}) dy - (e^x - e^{-x}) dx = 0

(e^x + e^{-x}) dy = (e^x - e^{-x}) dx

dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})} dx

Step 2: Integrate Both Sides
\int dy = \int \frac{(e^x - e^{-x})}{(e^x + e^{-x})} dx

Step 3: Perform the Integration
Let t = e^x + e^{-x} .
Then, differentiating with respect to x:
dt = (e^x - e^{-x}) dx .

This matches the numerator exactly. So the integral becomes:
\int dy = \int \frac{dt}{t}

y = \log|t| + C

Substitute t back:
y = \log|e^x + e^{-x}| + C

Final Answer: The general solution is \boxed{ y = \log(e^x + e^{-x}) + C } .

Q6: Find the general solution of the differential equation: \frac{dy}{dx} = (1+x^2)(1+y^2)

Solution:
Step 1: Separate the Variables
Group the y-terms with dy and x-terms with dx:
\frac{dy}{1+y^2} = (1+x^2) dx

Step 2: Integrate Both Sides
\int \frac{dy}{1+y^2} = \int (1+x^2) dx

Step 3: Perform the Integration

\tan^{-1} y = x + \frac{x^3}{3} + C

Final Answer: The general solution is \boxed{ \tan^{-1} y = x + \frac{x^3}{3} + C } .

Q7: Find the general solution of the differential equation: y \log y , dx - x , dy = 0

Solution:
Step 1: Separate the Variables
y \log y \ dx = x \ dy

\frac{dx}{x} = \frac{dy}{y \log y}

Step 2: Integrate Both Sides
\int \frac{dx}{x} = \int \frac{dy}{y \log y}

Step 3: Perform the Integration

• L.H.S.: \int \frac{1}{x} dx = \log |x|
• R.H.S.: Let t = \log y .
  Then, differentiating with respect to y, we get dt = \frac{1}{y} dy .
  The integral becomes \int \frac{1}{t} dt = \log |t| = \log |\log y| .

So, the equation becomes:
\log |x| = \log|\log y| + \log C

Step 4: Simplify the Solution Using logarithm properties
( \log m + \log n = \log mn )

So, the equation becomes:
\log |x| = \log|C \log y|

x = C \log y

Final Answer: The general solution is \boxed{ x = C \log y } (or alternatively y = e^{x/C} ).

Q8: Find the general solution of the differential equation: x^5 \frac{dy}{dx} = -y^5

Solution:
Step 1: Separate the Variables
x^5 \frac{dy}{dx} = -y^5

\frac{dy}{y^5} = - \frac{dx}{x^5}

Rewrite using negative exponents:
{y^{-5}} dy = - {x^{-5}} dx

Step 2: Integrate Both Sides
\int {y^{-5}} dy = - \int {x^{-5}} dx

\frac{y^{-4}}{-4} = - \frac{x^{-4}}{-4} + C

-1\frac{1}{4y^4} = \frac{1}{4x^4} + C

Step 3: Simplify the Expression
Multiply the entire equation by 4 to remove the fraction:

-1\frac{1}{y^4} = \frac{1}{x^4} + 4C

Rearrange terms:
-4C = \frac{1}{x^4} + 1\frac{1}{y^4}

Let -4C = k (an arbitrary constant).

x^{-4}} + y^{-4} = k

Final Answer: The general solution is \boxed{ x^{-4} + y^{-4} = k } .

Q9: Find the general solution of the differential equation: \frac{dy}{dx} = \sin^{-1} x

Solution:
Step 1: Separate the Variables
\frac{dy}{dx} = \sin^{-1} x

dy = \sin^{-1} x dx

Step 2: Integrate Both Sides
\int dy = \int \sin^{-1} x \ dx

y = \int \sin^{-1} x \ dx … (1)

Step 3: Perform the Integration
To integrate \sin^{-1} x , we use Integration by Parts.
Formula: \int u v , dx = u \int v , dx - \int (u' \int v , dx) , dx

Let u = \sin^{-1} x and v = 1 .
u = \sin^{-1} x \implies u' = \frac{1}{\sqrt{1-x^2}}

\int v , dx = \int 1 , dx = x

Applying the formula in (1):
y = x \sin^{-1} x - \int \left( \frac{1}{\sqrt{1-x^2}} \cdot x \right) dx

y = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx

Step 4: Solve the Second Integral
Let I_1 = \int \frac{x}{\sqrt{1-x^2}} dx .

Substitute 1 - x^2 = t^2 .
Then -2x \ dx = 2t \ dt \implies x \ dx = -t \ dt .
I_1 = \int \frac{-t \ dt}{t} = - \int dt = -t = -\sqrt{1-x^2}

Substitute I_1 back into the equation for y:
y = x \sin^{-1} x - (-\sqrt{1-x^2}) + C

y = x \sin^{-1} x + \sqrt{1-x^2} + C

Final Answer: The general solution is \boxed{ y = x \sin^{-1} x + \sqrt{1-x^2} + C } .

Q10: Find the general solution of the differential equation: e^x \tan y \ dx + (1 - e^x) \sec^2 y \ dy = 0

Solution:
Step 1: Separate the Variables Rearrange terms to separate x and y:
e^x \tan y \ dx = - (1 - e^x) \sec^2 y \ dy

Divide both sides by \tan y (1 - e^x) :
\frac{e^x}{1 - e^x} dx = - \frac{ \sec^2 y}{\tan y} \ dy

\frac{e^x}{e^x - 1} dx = \frac{ \sec^2 y}{\tan y} \ dy

Step 2: Integrate Both Sides
\int \frac{e^x}{e^x - 1} dx = \int \frac{ \sec^2 y}{\tan y} \ dy

Step 3: Perform the Integration

• L.H.S.: Let u = e^x - 1 ,
  then du = e^x dx .
  Integral is \int \frac{du}{u} = \log |u| = \log |e^x - 1| .
• R.H.S.: Let v = \tan y ,
  then dv = \sec^2 y , dy .
  Integral is \int \frac{dv}{v} = \log |v| = \log |\tan y| .

So, the equation becomes:
\log |e^x - 1| = \log |\tan y| + \log C

Step 4: Simplify
\log |e^x - 1| = \log |C \tan y|

e^x - 1 = C \tan y

\tan y = \frac{e^x - 1}{C}

\tan y = C(e^x - 1)

Final Answer: The general solution is \boxed{ \tan y = C(e^x - 1) } .

Q11: Find a particular solution of the differential equation satisfying the given condition: (x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \quad y = 1 \text{ when } x = 0

Solution:
(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x

Step 1: Separate the Variables First, factorize the term x^3 + x^2 + x + 1 :
x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1)

Now, write the differential equation in separated form:
(x^2+1)(x+1)\frac{dy}{dx} = 2x^2 + x

dy = \frac{2x^2 + x}{(x^2+1)(x+1)} dx … (1)

Step 2: Use Partial Fractions Decompose the R.H.S. into partial fractions:

\frac{2x^2 + x}{(x^2+1)(x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}

2x^2 + x = A(x^2+1) + (Bx+C)(x+1)

Find A: Put x = -1 :
2(-1)^2 + (-1) = A((-1)^2+1) \ implies 1 = 2A = \implies A = \frac{1}{2}

Find B and C: Compare coefficients.

Coefficient of x^2 : 2 = A + B \implies 2 = \frac{1}{2} + B \implies B = \frac{3}{2}

Constant term: 0 = A + C \implies 0 = \frac{1}{2} + C \implies C = -\frac{1}{2}

So, the fraction becomes:
\frac{2x^2 + x}{(x^2+1)(x+1)} = \frac{1}{2(x+1)} + \frac{\frac{3}{2}x - \frac{1}{2}}{x^2+1}

\frac{2x^2 + x}{(x^2+1)(x+1)} = \frac{1}{2(x+1)} + \frac{3x}{2(x^2+1)} - \frac{1}{2(x^2+1)}

Step 3: Integrate Both Sides
Substituting in (1) and integrating
\int dy = \frac{1}{2} \ int \frac{dx}{(x+1)} + \frac{3}{2} \frac{x}{(x^2+1)} \ dx - \frac{1}{2} \int \frac{dx}{(x^2+1)}

• \int \frac{dx}{x+1} = \log|x+1|
• \int \frac{x}{x^2+1} dx :
  Let t = x^2+1 , then dt=2x dx \implies x dx = dt/2 .
  Integral is \frac{1}{2} \log|x^2+1| .
• \int \frac{dx}{x^2+1} = \tan^{-1}x

The general solution is:
y = \frac{1}{2} \log|x+1| + \frac{3}{4} \log(x^2+1) - \frac{1}{2} \tan^{-1} x + C

Step 4: Apply Condition y=1, x=0
1 = \frac{1}{2} \log|1| + \frac{3}{4} \log(1) - \frac{1}{2} \tan^{-1} (0) + C

Since \log 1 = 0 and \tan^{-1} 0 = 0 :

1 = 0 + 0 - 0 + C \implies C = 1

y = \frac{1}{4} [2\log|x+1| + 3\log(x^2+1)] - \frac{1}{2} \tan^{-1}x + 1

y = \frac{1}{4} [\log|x+1|^2 + \log(x^2+1)^3] - \frac{1}{2} \tan^{-1}x + 1

y = \frac{1}{4} \log[(x+1)^2 (x^2+1)^3] - \frac{1}{2} \tan^{-1}x + 1

Final Answer: The particular solution is \boxed{ y = \frac{1}{4} \log[(x+1)^2 (x^2+1)^3] - \frac{1}{2} \tan^{-1}x + 1 } .

Q12: Find a particular solution of the differential equation satisfying the given condition: x(x^2 - 1)\frac{dy}{dx} = 1; \quad y = 0 \text{ when } x = 2

Solution:
Step 1: Separate the Variables
x(x^2 - 1)\frac{dy}{dx} = 1

dy = \frac{dx}{x(x^2-1)}

dy = \frac{dx}{x(x-1)(x+1)}

Step 2: Use Partial Fractions
\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}

1 = A(x^2-1) + Bx(x+1) + Cx(x-1)

• Put x = 0 \implies 1 = A(-1) \implies A = -1
• Put x = 1 \implies 1 = B(1)(2) \implies B = \frac{1}{2}
• Put x = -1 \implies 1 = C(-1)(-2) \implies C = \frac{1}{2}

Step 3: Integrate Both Sides

y = -\log|x| + \frac{1}{2} \log|x-1| + \frac{1}{2} \log|x+1| + C

Simplify using log properties:

y = -\log|x| + \frac{1}{2} \log|x^2-1| + C

Step 4: Apply Condition y=0, x=2

0 = -\log 2 + \frac{1}{2} \log 3 + C

Final Answer: The particular solution is \boxed{ y = \frac{1}{2} \log|x^2-1| - \log|x| + \log \left( \frac{2}{\sqrt{3}} \right) } .

Q13: Find a particular solution of the differential equation satisfying the given condition: \cos(\frac{dy}{dx}) = a , (a \in \mathbb{R}); \quad y = 1 \text{ when } x = 0

Solution:
Step 1: Simplify the Differential Equation
\cos(\frac{dy}{dx}) = a

\frac{dy}{dx} = \cos^{-1} a

Since a is a constant, \cos^{-1} a is also a constant.

Step 2: Separate and Integrate
dy = (\cos^{-1} a ) dx

\int dy = \int (\cos^{-1} a) dx

y = x (\cos^{-1} a ) + C

Step 3: Apply Condition y=1, x=0
1 = 0 \cdot (\cos^{-1} a) + C \implies C = 1

Step 4: Substitute C back

y = x (\cos^{-1} a ) + 1

Rearrange to match standard form:
y - 1 = x (\cos^{-1} a )

\frac{y - 1}{x} = x (\cos^{-1} a )

Final Answer: The particular solution is \boxed{ \cos \left( \frac{y-1}{x} \right) = a }

Q14: Find a particular solution of the differential equation satisfying the given condition: \frac{dy}{dx} = y \tan x; \quad y = 1 \text{ when } x = 0

Solution:
Step 1: Separate the Variables
\frac{dy}{dx} = y \tan x

\frac{dy}{y} = \tan x \ dx

Step 2: Integrate Both Sides
\int \frac{dy}{y} = \int \tan x \ dx

\log|y| = \log|\sec x| + C

Step 3: Apply Condition y=1, x=0
\log|1| = \log|\sec 0| + C

0 = \log(1) + C \implies 0 = 0 + C \implies C = 0

Step 4: Substitute C back
\log|y| = \log|\sec x|

y = \sec x

Final Answer: The particular solution is \boxed{ y = \sec x } .

Q15: Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x \sin x .

Solution:

Step 1: Separate the Variables
The given differential equation is:

\frac{dy}{dx} = e^x \sin x

Separate the variables: dy = e^x \sin x \ dx

Step 2: Integrate Both Sides
\int dy = \int e^x \sin x \ dx

y = \int e^x \sin x , dx

Let I = \int e^x \sin x , dx .

We use Integration by Parts where \int u v' dx = uv - \int u'v dx .
Let u = \sin x and v' = e^x .
Then u' = \cos x and v = e^x .

I = \sin x \cdot e^x - \int \cos x \cdot e^x \ dx … (1)

Now, apply integration by parts again for \int e^x \cos x , dx .
Let u = \cos x and v' = e^x .
Then u' = -\sin x and v = e^x .

\int e^x \cos x \ dx = \cos x \cdot e^x - \int (-\sin x) e^x \ dx = e^x \cos x + \int e^x \sin x \ dx

\int e^x \cos x \ dx = e^x \cos x + I

Substitute this back into the equation (1) for I :
I = \sin x \cdot e^x - (e^x \cos x + I)

I = e^x \sin x - e^x \cos x - I

2I = e^x (\sin x - \cos x)

I = \frac{e^x}{2}(\sin x - \cos x)

So, the general solution is:
y = \frac{e^x}{2}(\sin x - \cos x) + C

Step 3: Apply the Condition (0, 0)
Substitute x = 0 and y = 0 :
0 = \frac{e^0}{2}(\sin 0 - \cos 0) + C

0 = \frac{1}{2}(0 - 1) + C

0 = -\frac{1}{2} + C \implies C = \frac{1}{2}

Step 4: Substitute C back
y = \frac{e^x}{2}(\sin x - \cos x) + \frac{1}{2}

2y = e^x(\sin x - \cos x) + 1

Final Answer: The equation of the curve is \boxed{ 2y = e^x(\sin x - \cos x) + 1 } .

Q16: For the differential equation xy \frac{dy}{dx} = (x+2)(y+2) , find the solution curve passing through the point (1, -1) .

Solution:
Step 1: Separate the Variables
xy \frac{dy}{dx} = (x+2)(y+2)

\frac{y}{y+2} dy = \frac{x+2}{x} dx

Step 2: Simplify Integrands:
L.H.S.: \frac{y}{y+2} = \frac{y+2-2}{y+2} = 1 - \frac{2}{y+2}

R.H.S.: \frac{x+2}{x} = 1 + \frac{2}{x}

Step 3: Integrate Both Sides
\int (1 - \frac{2}{y+2}) dy = \int( 1 + \frac{2}{x}) \ dx

y - 2 \log|y+2| = x + 2 \log|x| + C

Step 4: Apply Condition (1, -1)
Substitute x = 1 and y = -1 :
-1 - 2 \log|-1+2| = x + 2 \log|1| + C
-1 - 2 \log(1) = 1 + 2(0) + C
-1 - 0 = 1 + 0 + C
C = -2

Step 5: Substitute C back and Simplify
y - 2 \log|y+2| = x + 2 \log|x| - 2
y - x + 2 = 2 \log|x| + 2 \log|y+2|
y - x + 2 = 2 (\log|x| + \log|y+2|)
y - x + 2 = 2 (\log|x(y+2)|)
y - x + 2 = \log|x(y+2)|^2
y - x + 2 = \log(x^2(y+2)^2)

Final Answer: The solution curve is \boxed{ y - x + 2 = \log(x^2(y+2)^2) } .

Q17: Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

Solution:
Step 1: Formulate the Differential Equation

Slope of the tangent at (x, y) is \frac{dy}{dx} .

The problem states: Product of slope and y-coordinate = x-coordinate.
y \cdot \frac{dy}{dx} = x

Step 2: Separate the Variables
y \ dy = x \ dx

Step 3: Integrate Both Sides
\int y \ dy = \int x \ dx

\frac{y^2}{2} = \frac{x^2}{2} + C

y^2 = x^2 + 2C
y^2 - x^2 = C'

Step 4: Apply Condition (0, -2)
Substitute x = 0 and y = -2 :

(-2)^2 - 0^2 = C'

4 = C'

Step 5: Substitute C’ back
y^2 - x^2 = 4

Final Answer: The equation of the curve is \boxed{ y^2 - x^2 = 4 } (This is a equilateral hyperbola).

Q18: At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3) . Find the equation of the curve given that it passes through (-2, 1) .

Solution:
Step 1: Formulate the Differential Equation
Slope of the tangent at (x, y) is m_1 = \frac{dy}{dx} .
Slope of the line segment joining (x, y) and (-4, -3) is:
m_2 = \frac{y - (-3)}{x - (-4)} = \frac{y+3}{x+4}

The condition is given as m_1 = 2 m_2 :
\frac{dy}{dx} = 2 \left( \frac{y+3}{x+4} \right)

Step 2: Separate the Variables
\frac{dy}{y+3} = \frac{2, dx}{x+4}

Step 3: Integrate Both Sides
\int \frac{dy}{y+3} = \int \frac{2, dx}{x+4}

\log|y+3| = 2 \log|x+4| + \log C

Step 4: Simplify Using Log Properties
\log|y+3| = \log|(x+4)^2| + \log C

\log|y+3| = \log|C(x+4)^2|

y+3 = C(x+4)^2

Step 5: Apply Condition (-2, 1)
Substitute x = -2 and y = 1 :
1+3 = C(-2+4)^2
4 = C(2)^2
4 = 4C \implies C = 1

Step 6: Substitute C back
y+3 = 1(x+4)^2

Final Answer: The equation of the curve is \boxed{ y + 3 = (x+4)^2 } .

Q19: The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of balloon after t seconds.

Solution:
Step 1: Formulate the Differential Equation
Let V be the volume and r be the radius of the balloon at time t .
Volume of a sphere: V = \frac{4}{3} \pi r^3 .

The problem states the rate of change of volume is constant:
\frac{dV}{dt} = k (where k is a constant).

\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3} \pi r^3) = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} .

So, the differential equation is:
4\pi r^2 \frac{dr}{dt} = k

Step 2: Separate and Integrate
4\pi r^2 \ dr = k \ dt

Integrate both sides:
\int 4\pi r^2 \ dr = \int k \ dt

4\pi \frac{r^3}{3} = kt + C … (1)

Step 3: Apply Initial Condition ( t=0, r=3 )
Substitute t=0 and r=3 into (1):

\frac{4 \pi}{3} 3^3 = k(0) + C

\frac{4\pi}{3} (27) = C

36\pi = C

Substitute C back into (1):
\frac{4\pi r^3}{3} = kt + 36\pi … (2)

Step 4: Apply Second Condition ( t=3, r=6 )
Substitute t=3 and r=6 into (2):

\frac{4\pi \cdot 6^3}{3} = k(3) + 36\pi

\frac{4\pi}{3} (216) = 3k + 36\pi

252 \pi = 3k \implies k = 84 \pi

Step 5: Find the Final Equation
Substitute k back into equation (2):

\frac{4\pi r^3}{3} = 84 \pi t + 36\pi

\frac{r^3}{3} = 21 t + 9

r^3 = 63 t + 27
Take the cube root:
r = (63 t + 27)^{\frac{1}{3}}

Final Answer: The radius of the balloon after t seconds is \boxed{ r = (63t + 27)^{\frac{1}{3}} } .

Q20: In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years ( \log_e 2 = 0.6931 ).

Solution:
Step 1: Formulate the Differential Equation
Let P be the principal at time t .
The rate of change is proportional to P :
\frac{dP}{dt} = \frac{r}{100} \times P

Step 2: Separate and Integrate
\frac{dP}{P} = \frac{r}{100} dt

Integrate both sides:
\int \frac{dP}{P} = \frac{r}{100} \int dt

\log P = \frac{rt}{100} + C … (1)

Step 3: Apply Initial Condition
Let initial principal be P_0 . At t=0, P=P_0 .
\log P_0 = 0 + C \implies C = \log P_0

Substitute back into (1):
\log P = \frac{rt}{100} + \log P_0

\log P - \log P_0 = \frac{rt}{100}

\log (\frac{P}{P_0}) = \frac{rt}{100} … (2)

Step 4: Apply Given Condition (Doubling)
Given P_0 = 100 and P = 200 (doubles) when t = 10 .
Substitute into (2):

\log (\frac{200}{100}) = \frac{r(10)}{100}

\log 2 = \frac{r}{10}

Given \log_e 2 = 0.6931 :
0.6931 = \frac{r}{10}

r = 6.931

Final Answer: The value of r is \boxed{ \text{6.93 %} } .

Q21: In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years ( e^{0.5} = 1.648 ).

Solution:

Step 1: Use the General Formula
From the previous question, we know the solution for continuous growth is
\log (\frac{P}{P_0}) = \frac{rt}{100} or P = P_0 e^{\frac{rt}{100}} .

Step 2: Substitute Known Values
Initial Principal P_0 = 1000
Rate r = 5
Time t = 10

P = 1000 e^{\frac{5 \times 10}{100}}

P = 1000 e^{\frac{50}{100}}

P = 1000 e^{0.5}

Step 3: Calculate Final Amount
Given e^{0.5} = 1.648 :
P = 1000 \times 1.648
P = 1648

Final Answer: The amount will be worth \boxed{ \text{Rs } 1648 } .

Q22: In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution:
Step 1: Formulate the Differential Equation
Let N be the number of bacteria at time t .
\frac{dN}{dt} \propto N \implies \frac{dN}{dt} = kN .

Solving this (Variables Separable):
\frac{dN}{N} = k \ dt .

\log N = kt + C

N = C_1 e^{kt} (where C_1 = e^C ).

Step 2: Find the Constant C
At t=0 , N = 1,00,000 .
1,00,000 = C_1 e^0 \implies C_1 = 1,00,000 .

So, the equation is: N = 1,00,000 e^{kt} … (1)

Step 3: Find the Rate Constant k
At t=2 , the number increases by 10%.
New N = 1,00,000 + 10% \text{ of } 1,00,000 = 1,10,000 .
1,10,000 = 1,00,000 e^{k(2)}

Substituting into (1):
\frac{11}{10} = e^{2k}

1.1 = e^{2k}

Taking log: \log (1.1) = 2k \implies k = \frac{1}{2} \log (1.1) .

Step 4: Find Time t for N = 2,00,000
Substitute N = 2,00,000 and the value of k into (1):
2,00,000 = 1,00,000 e^{kt}
2 = e^{kt}
\log 2 = kt

t = \frac{\log 2}{k}

Substitute k = \frac{1}{2} \log (1.1) :

t = \frac{2\log 2}{\log (1.1)}

Final Answer: The count will reach 2,00,000 in \boxed{ \frac{2 \log 2}{\log 1.1} } hours.

Q23: The general solution of the differential equation \frac{dy}{dx} = e^{x+y} is:

(A) e^x + e^{-y} = C

(B) e^x + e^y = C

(C) e^{-x} + e^y = C

(D) e^{-x} + e^{-y} = C

Solution:
Step 1: Separate the Variables
\frac{dy}{dx} = e^{x+y}

\frac{dy}{dx} = e^x \cdot e^y

\frac{dy}{e^y} = e^x \ dx

e^{-y} \ dy = e^x \ dx

Step 2: Integrate Both Sides
\int e^{-y} \ dy = \int e^x \ dx

\frac{e^{-y}}{-1} = e^x + K

-e^{-y} = e^x + K

Step 3: Rearrange to Match Options
e^x + e^{-y} = -K

e^x + e^{-y} = C

Final Answer: The correct option is (A) .