Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations Exercise 4.5

Question 1. \( \frac { d^2y }{dx^2} – 6\frac { dy }{dx} + 8y = 0 \)
Answer: Given the differential equation \( (D^2 – 6D + 8) y = 0 \), where \( D = \frac{d}{dx} \).
First, we form the auxiliary equation, which helps us find the general solution.
\( m^2 – 6m + 8 = 0 \)
Factoring this quadratic equation gives:
\( (m – 4)(m - 2) = 0 \)
This means the roots are \( m = 4 \) and \( m = 2 \). Since these roots are real and different, we can write the complementary function (C.F).
The complementary function (C.F) is \( (Ae^{4x} + Be^{2x}) \).
The general solution for this differential equation is \( y = Ae^{4x} + Be^{2x} \).
In simple words: We find values for 'm' that solve a simpler equation related to the problem. Since these 'm' values are real and not the same, we use them to build the main answer for 'y', which includes two different exponential terms with constants A and B.

🎯 Exam Tip: Always identify the type of roots (real and distinct, real and equal, or complex) as this determines the specific form of the complementary function.

Question 2. \( \frac { d^2y }{dx^2} – 4\frac { dy }{dx} + 4y = 0 \)
Answer: Given the differential equation \( \frac{d^2y}{dx^2} – 4\frac{dy}{dx} + 4y = 0 \).
The auxiliary equation (A.E) is formed from the coefficients of the derivatives.
\( m^2 – 4m + 4 = 0 \)
This is a perfect square and can be factored as:
\( (m - 2)^2 = 0 \)
So, the roots are \( m = 2, 2 \). Here, the roots are real and equal.
For real and equal roots, the complementary function (C.F) takes a specific form.
The complementary function (C.F) is \( (Ax + B) e^{2x} \).
Therefore, the general solution for this differential equation is \( y = (Ax + B) e^{2x} \).
In simple words: We convert the problem into a simple quadratic equation. If the solutions to this equation are the same number, we use that number along with an 'x' term and constants A and B to write the final answer.

🎯 Exam Tip: When roots are real and equal, remember to multiply the 'A' constant by 'x' in the complementary function.

Question 3. \( (D^2 + 2D + 3) y = 0 \)
Answer: Given the differential equation \( (D^2 + 2D + 3) y = 0 \).
We form the auxiliary equation from the given differential operator.
The auxiliary equation is \( m^2 + 2m + 3 = 0 \).
Here, comparing with \( am^2 + bm + c = 0 \), we have \( a = 1, b = 2, c = 3 \).
We use the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots.
\( m = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(3)}}{2(1)} \)

\( \implies m = \frac{-2 \pm \sqrt{4 - 12}}{2} \)

\( \implies m = \frac{-2 \pm \sqrt{-8}}{2} \)
We can rewrite \( \sqrt{-8} \) as \( \sqrt{4 \times -2} = 2\sqrt{-2} = 2\sqrt{2}i \).
So, \( m = \frac{-2 \pm 2\sqrt{2}i}{2} \)

\( \implies m = -1 \pm \sqrt{2}i \)
Here, the roots are complex, of the form \( \alpha \pm i\beta \), where \( \alpha = -1 \) and \( \beta = \sqrt{2} \).
The complementary function (C.F) for complex roots is \( e^{\alpha x} (A\cos\beta x + B\sin\beta x) \).
Therefore, the complementary function is \( e^{-x} [A\cos\sqrt{2}x + B\sin\sqrt{2}x] \).
This also forms the general solution, so \( y = e^{-x} (A\cos\sqrt{2}x + B\sin\sqrt{2}x) \).
In simple words: When we solve the simple equation for 'm', sometimes the answers involve imaginary numbers. In this case, we use those imaginary numbers to create a solution for 'y' that involves 'e' to a power and sine and cosine functions. This form helps describe oscillations.

🎯 Exam Tip: When using the quadratic formula, pay close attention to the sign under the square root, as a negative sign indicates complex roots requiring the \( e^{\alpha x}(A\cos\beta x + B\sin\beta x) \) form.

Question 4. \( \frac { d^2y }{dx^2} – 2k\frac { dy }{dx} + k^2y = 0 \)
Answer: Given the differential equation \( \frac{d^2y}{dx^2} – 2k\frac{dy}{dx} + k^2y = 0 \).
We write this in operator form as \( (D^2 – 2kD + k^2)y = 0 \), where \( D = \frac{d}{dx} \).
The auxiliary equation is \( m^2 – 2km + k^2 = 0 \).
This equation is a perfect square, just like in an algebraic identity.
\( (m – k)^2 = 0 \)
So, the roots are \( m = k, k \). These are real and equal roots.
For real and equal roots, the complementary function (C.F) has a specific structure.
The complementary function (C.F) is \( (Ax + B) e^{kx} \).
The general solution for this differential equation is \( y = (Ax + B) e^{kx} \).
In simple words: We turn the problem into a simple equation, and if the solution 'm' is the same number 'k' twice, then the final answer 'y' will be based on 'e' to the power of 'kx', multiplied by an expression with 'x' and constants. This shows how repeated roots affect the solution.

🎯 Exam Tip: Recognizing perfect squares in auxiliary equations saves time and helps identify real and equal roots quickly.

Question 5. \( (D^2 – 2D – 15) y = 0 \)
Answer: Given the differential equation \( (D^2 – 2D – 15) y = 0 \).
We first find the auxiliary equation.
The auxiliary equation is \( m^2 - 2m - 15 = 0 \).
We factor the quadratic equation to find its roots:
\( m^2 + 3m – 5m – 15 = 0 \)
\( m (m + 3) – 5 (m + 3) = 0 \)
\( (m + 3) (m - 5) = 0 \)
So, the roots are \( m = -3, 5 \). These roots are real and different.
The complementary function (C.F) for real and distinct roots is \( Ae^{m_1 x} + Be^{m_2 x} \).
Substituting the roots, the C.F. becomes \( Ae^{-3x} + Be^{5x} \).
The general solution for this homogeneous differential equation is \( y = Ae^{-3x} + Be^{5x} \). (1)
Now, we differentiate the general solution to find \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = Ae^{-3x}(-3) + Be^{5x}(5) \)
\( \frac{dy}{dx} = -3Ae^{-3x} + 5Be^{5x} \) (2)
Next, we differentiate again to find \( \frac{d^2y}{dx^2} \).
\( \frac{d^2y}{dx^2} = -3A(-3)e^{-3x} + 5B(5)e^{5x} \)
\( \frac{d^2y}{dx^2} = 9Ae^{-3x} + 25Be^{5x} \) (3)
The problem statement is incomplete at this point, but it appears to be implicitly setting some conditions based on further steps in the original source, assuming conditions like \( \frac{dy}{dx} = 0 \) and \( \frac{d^2y}{dx^2} = 0 \) when \( x = 0 \). Let's follow the provided steps for the values of A and B.
If we assume the condition is that \( \frac{dy}{dx} = 0 \) when \( x = 0 \):
Substitute \( x = 0 \) and \( \frac{dy}{dx} = 0 \) into equation (2):
\( 0 = -3Ae^0 + 5Be^0 \)
\( -3A + 5B = 0 \) (4)
If we assume the condition is that \( \frac{d^2y}{dx^2} = 2 \) when \( x = 0 \):
Substitute \( x = 0 \) and \( \frac{d^2y}{dx^2} = 2 \) into equation (3):
\( 2 = 9Ae^0 + 25Be^0 \)
\( 9A + 25B = 2 \) (5)
Now, we solve equations (4) and (5) simultaneously for A and B.
Multiply equation (4) by 3:
\( 3(-3A + 5B) = 3(0) \)
\( -9A + 15B = 0 \) (Eqn (4) × 3)
Now add this to equation (5):
\( (9A + 25B) + (-9A + 15B) = 2 + 0 \)
\( 40B = 2 \)
\( B = \frac{2}{40} \)
\( B = \frac{1}{20} \)
Substitute \( B = \frac{1}{20} \) into equation (4):
\( -3A + 5(\frac{1}{20}) = 0 \)
\( -3A + \frac{1}{4} = 0 \)
\( -3A = -\frac{1}{4} \)
\( A = \frac{1}{12} \)
Substitute the values of \( A = \frac{1}{12} \) and \( B = \frac{1}{20} \) back into the general solution (1).
So, \( y = \frac{1}{12}e^{-3x} + \frac{1}{20}e^{5x} \).
In simple words: We find the 'm' values from the main equation. These tell us the basic form of the answer. Then, using extra clues (like the value of 'y' or its changes at a specific point), we figure out the exact numbers (A and B) to make the solution complete.

🎯 Exam Tip: For problems involving initial conditions, remember to differentiate the general solution and substitute the conditions to solve for the constants A and B.

Question 6. \( (4D^2 + 4D – 3) y = e^{2x} \)
Answer: Given the differential equation \( (4D^2 + 4D – 3) y = e^{2x} \).
First, we find the auxiliary equation to determine the complementary function.
The auxiliary equation is \( 4m^2 + 4m - 3 = 0 \).
We factor this quadratic equation:
\( 4m^2 + 6m – 2m - 3 = 0 \)
\( 2m (2m + 3) – 1 (2m + 3) = 0 \)
\( (2m + 3) (2m – 1) = 0 \)
This gives us two distinct roots:
\( 2m = -3 \implies m = -\frac{3}{2} \)
\( 2m = 1 \implies m = \frac{1}{2} \)
Since the roots are real and different, the complementary function is \( C.F = Ae^{\frac{1}{2}x} + Be^{-\frac{3}{2}x} \).
Next, we find the particular integral (P.I) for the non-homogeneous part \( e^{2x} \).
\( P.I = \frac{1}{4D^2 + 4D - 3} e^{2x} \)
Substitute \( D = 2 \) (from the exponent of \( e^{2x} \)) into the denominator:
\( P.I = \frac{1}{4(2)^2 + 4(2) - 3} e^{2x} \)
\( P.I = \frac{1}{4(4) + 8 - 3} e^{2x} \)
\( P.I = \frac{1}{16 + 8 - 3} e^{2x} \)
\( P.I = \frac{1}{21} e^{2x} \)
The general solution is the sum of the complementary function and the particular integral.
The general solution is \( y = C.F + P.I \).
So, \( y = Ae^{\frac{1}{2}x} + Be^{-\frac{3}{2}x} + \frac{e^{2x}}{21} \).
In simple words: We find two parts of the answer: one for the core equation (C.F) and another for the extra term (\( e^{2x} \)) on the right side (P.I). We combine these two parts to get the full answer. The C.F. deals with the natural behavior, and the P.I. with the forced behavior.

🎯 Exam Tip: Remember to calculate both the complementary function and the particular integral for non-homogeneous differential equations, and ensure the denominator does not become zero when finding P.I.

Question 7. \( \frac { d^2y }{dx^2} + 16y = 0 \)
Answer: Given the differential equation \( \frac{d^2y}{dx^2} + 16y = 0 \).
We can write this in operator form as \( (D^2 + 16) y = 0 \).
The auxiliary equation is \( m^2 + 16 = 0 \).
Solving for m:
\( m^2 = -16 \)
\( m = \pm \sqrt{-16} \)
\( m = \pm 4i \)
These are complex roots of the form \( \alpha \pm i\beta \), where \( \alpha = 0 \) (since there is no real part) and \( \beta = 4 \).
The complementary function (C.F) for complex roots is \( e^{\alpha x} (A\cos\beta x + B\sin\beta x) \).
Substituting the values, the C.F. is \( e^{0x} [A\cos 4x + B\sin 4x] \).
Since \( e^{0x} = 1 \), the complementary function simplifies to \( [A\cos 4x + B\sin 4x] \).
As this is a homogeneous equation, the general solution is simply the complementary function.
The general solution is \( y = [A\cos 4x + B\sin 4x] \).
In simple words: When the equation leads to imaginary numbers as solutions, the final answer will involve cosine and sine waves. This form of solution often describes things that vibrate or swing back and forth, like a pendulum.

🎯 Exam Tip: If the auxiliary equation results in purely imaginary roots (like \( \pm bi \)), the real part \( \alpha \) is 0, making the \( e^{\alpha x} \) term equal to 1.

Question 8. \( (D^2 – 3D + 2) y = e^{3x} \) which shall vanish for \( x = 0 \) and for \( x = \log 2 \)
Answer: Given the differential equation \( (D^2 – 3D + 2) y = e^{3x} \).
First, we find the auxiliary equation.
The auxiliary equation is \( m^2 – 3m + 2 = 0 \).
Factoring the quadratic equation:
\( (m - 1) (m - 2) = 0 \)
So, the roots are \( m = 1, 2 \). These are real and different roots.
The complementary function (C.F) is \( Ae^{m_1 x} + Be^{m_2 x} \).
Therefore, the C.F. is \( Ae^{x} + Be^{2x} \).
Next, we find the particular integral (P.I) for \( e^{3x} \).
\( P.I = \frac{1}{D^2 - 3D + 2} e^{3x} \)
Substitute \( D = 3 \) (from the exponent of \( e^{3x} \)) into the denominator.
\( P.I = \frac{1}{(3)^2 - 3(3) + 2} e^{3x} \)
\( P.I = \frac{1}{9 - 9 + 2} e^{3x} \)
\( P.I = \frac{1}{2} e^{3x} \)
The general solution is \( y = C.F + P.I \).
So, \( y = Ae^{x} + Be^{2x} + \frac{e^{3x}}{2} \) (1)
Now, we use the given conditions. The solution "vanishes" (meaning \( y = 0 \)) at two points.
Condition 1: When \( x = 0, y = 0 \).
Substitute these into equation (1):
\( 0 = Ae^0 + Be^0 + \frac{e^0}{2} \)
\( 0 = A + B + \frac{1}{2} \)
\( A + B = -\frac{1}{2} \) (2)
Condition 2: When \( x = \log 2, y = 0 \).
Substitute these into equation (1):
\( 0 = Ae^{\log 2} + Be^{2\log 2} + \frac{e^{3\log 2}}{2} \)
Using properties of logarithms \( e^{\log a} = a \) and \( e^{n\log a} = e^{\log a^n} = a^n \):
\( 0 = A(2) + B(2^2) + \frac{(2^3)}{2} \)
\( 0 = 2A + 4B + \frac{8}{2} \)
\( 0 = 2A + 4B + 4 \)
\( 2A + 4B = -4 \) (3)
Now, we solve equations (2) and (3) simultaneously for A and B.
Multiply equation (2) by 2:
\( 2(A + B) = 2(-\frac{1}{2}) \)
\( 2A + 2B = -1 \) (Eqn (2) × 2)
Subtract this new equation from equation (3):
\( (2A + 4B) - (2A + 2B) = -4 - (-1) \)
\( 2B = -3 \)
\( B = -\frac{3}{2} \)
Substitute \( B = -\frac{3}{2} \) into equation (2):
\( A + (-\frac{3}{2}) = -\frac{1}{2} \)
\( A = -\frac{1}{2} + \frac{3}{2} \)
\( A = \frac{2}{2} \)
\( A = 1 \)
Finally, substitute the values of A and B back into the general solution (1).
The specific solution is \( y = 1e^{x} + (-\frac{3}{2})e^{2x} + \frac{e^{3x}}{2} \).
So, \( y = e^{x} - \frac{3}{2}e^{2x} + \frac{e^{3x}}{2} \).
In simple words: We first find a basic solution, then use the extra clues (that the answer is zero at two specific points) to find the exact numbers for the constants in our solution. This gives us one unique answer that fits all the given information.

🎯 Exam Tip: Pay close attention to initial/boundary conditions. They are crucial for determining the unique values of constants A and B in the general solution.

Question 9. \( (D^2 + D - 2) y = e^{3x} + e^{-3x} \)
Answer: Given the differential equation \( (D^2 + D - 2) y = e^{3x} + e^{-3x} \).
First, we find the auxiliary equation.
The auxiliary equation is \( m^2 + m - 2 = 0 \).
Factoring the quadratic equation:
\( (m + 2) (m - 1) = 0 \)
So, the roots are \( m = -2, 1 \). These are real and different roots.
The complementary function (C.F) is \( Ae^{m_1 x} + Be^{m_2 x} \).
Therefore, the C.F. is \( Ae^{-2x} + Be^{x} \).
Next, we find the particular integral (P.I). Since the right-hand side has two terms, we find two particular integrals, \( P.I_1 \) for \( e^{3x} \) and \( P.I_2 \) for \( e^{-3x} \).
For \( P.I_1 \):
\( P.I_1 = \frac{1}{D^2 + D - 2} e^{3x} \)
Substitute \( D = 3 \):
\( P.I_1 = \frac{1}{(3)^2 + (3) - 2} e^{3x} \)
\( P.I_1 = \frac{1}{9 + 3 - 2} e^{3x} \)
\( P.I_1 = \frac{1}{10} e^{3x} \)
For \( P.I_2 \):
\( P.I_2 = \frac{1}{D^2 + D - 2} e^{-3x} \)
Substitute \( D = -3 \). First, check if the denominator becomes zero:
\( (-3)^2 + (-3) - 2 = 9 - 3 - 2 = 4 \). It's not zero, so we can directly substitute.
\( P.I_2 = \frac{1}{(-3)^2 + (-3) - 2} e^{-3x} \)
\( P.I_2 = \frac{1}{9 - 3 - 2} e^{-3x} \)
\( P.I_2 = \frac{1}{4} e^{-3x} \)
The total particular integral is \( P.I = P.I_1 + P.I_2 \).
\( P.I = \frac{1}{10} e^{3x} + \frac{1}{4} e^{-3x} \)
The general solution is \( y = C.F + P.I \).
So, \( y = Ae^{-2x} + Be^{x} + \frac{1}{10} e^{3x} + \frac{1}{4} e^{-3x} \).
In simple words: We find the basic answer part (C.F) from the left side of the equation. Then, because the right side has two separate exponential terms, we find a specific answer for each term (P.I.s) and add them all together to get the full solution.

🎯 Exam Tip: When the right-hand side is a sum of terms, find the particular integral for each term separately and then add them up. If a particular integral causes the denominator to be zero, use the special case rule (multiply by x and differentiate the denominator).

Question 10. \( (D^2 – 10D + 25) y = 4e^{5x} + 5 \)
Answer: Given the differential equation \( (D^2 – 10D + 25) y = 4e^{5x} + 5 \).
First, find the auxiliary equation for the complementary function.
The auxiliary equation is \( m^2 – 10m + 25 = 0 \).
This is a perfect square:
\( (m – 5) (m – 5) = 0 \)
So, the roots are \( m = 5, 5 \). These are real and equal roots.
The complementary function (C.F) is \( (Ax + B) e^{mx} \).
Therefore, the C.F. is \( (Ax + B) e^{5x} \).
Next, we find the particular integral for each term on the right-hand side. Let \( P.I_1 \) be for \( 4e^{5x} \) and \( P.I_2 \) for \( 5 \).
For \( P.I_1 \):
\( P.I_1 = \frac{1}{D^2 - 10D + 25} (4e^{5x}) \)
Here, substituting \( D = 5 \) makes the denominator zero: \( (5)^2 - 10(5) + 25 = 25 - 50 + 25 = 0 \).
So, we use the special case: multiply by \( x \) and differentiate the denominator with respect to \( D \).
\( P.I_1 = x \frac{1}{2D - 10} (4e^{5x}) \)
Again, substituting \( D = 5 \) makes \( 2(5) - 10 = 0 \). We repeat the process.
\( P.I_1 = x^2 \frac{1}{2} (4e^{5x}) \)
\( P.I_1 = x^2 (2e^{5x}) \)
\( P.I_1 = 2x^2 e^{5x} \).
For \( P.I_2 \):
\( P.I_2 = \frac{1}{D^2 - 10D + 25} (5) \)
We can write \( 5 \) as \( 5e^{0x} \). Substitute \( D = 0 \).
\( P.I_2 = \frac{1}{(0)^2 - 10(0) + 25} (5) \)
\( P.I_2 = \frac{5}{25} \)
\( P.I_2 = \frac{1}{5} \)
The general solution is \( y = C.F + P.I_1 + P.I_2 \).
So, \( y = (Ax + B) e^{5x} + 2x^2 e^{5x} + \frac{1}{5} \).
In simple words: We find the first part of the answer using the equation's left side. Then, for the right side, because it has two parts, we find a specific answer for each part separately. One part required an extra step because a simple replacement would make the denominator zero. Finally, we add all these parts to get the complete solution.

🎯 Exam Tip: When calculating the particular integral, always check if substituting D directly makes the denominator zero. If it does, remember to use the special rule of multiplying by x and differentiating the denominator, repeating if necessary.

Question 11. \( (4D^2 + 16D +15) y = 4e^{-\frac{3}{2}x} \)
Answer: Given the differential equation \( (4D^2 + 16D +15) y = 4e^{-\frac{3}{2}x} \).
First, find the auxiliary equation.
The auxiliary equation is \( 4m^2 + 16m + 15 = 0 \).
Factor this quadratic equation:
\( 4m^2 + 10m + 6m + 15 = 0 \)
\( 2m (2m + 5) + 3 (2m + 5) = 0 \)
\( (2m + 3) (2m + 5) = 0 \)
This gives us two distinct roots:
\( 2m = -3 \implies m = -\frac{3}{2} \)
\( 2m = -5 \implies m = -\frac{5}{2} \)
Since the roots are real and different, the complementary function (C.F) is \( Ae^{-\frac{3}{2}x} + Be^{-\frac{5}{2}x} \).
Next, we find the particular integral (P.I) for \( 4e^{-\frac{3}{2}x} \).
\( P.I = \frac{1}{4D^2 + 16D + 15} (4e^{-\frac{3}{2}x}) \)
Substitute \( D = -\frac{3}{2} \) (from the exponent). Check if the denominator is zero:
\( 4(-\frac{3}{2})^2 + 16(-\frac{3}{2}) + 15 = 4(\frac{9}{4}) - 24 + 15 = 9 - 24 + 15 = 0 \).
Since the denominator is zero, we must use the special rule: multiply by \( x \) and differentiate the denominator with respect to \( D \).
\( P.I = x \frac{1}{8D + 16} (4e^{-\frac{3}{2}x}) \)
Now, substitute \( D = -\frac{3}{2} \) again into the new denominator:
\( 8(-\frac{3}{2}) + 16 = -12 + 16 = 4 \). This is not zero.
So, \( P.I = x \frac{1}{4} (4e^{-\frac{3}{2}x}) \)
\( P.I = x e^{-\frac{3}{2}x} \).
The general solution is \( y = C.F + P.I \).
So, \( y = Ae^{-\frac{3}{2}x} + Be^{-\frac{5}{2}x} + xe^{-\frac{3}{2}x} \).
In simple words: We find the basic solution (C.F) from the left side of the equation. For the right side, because replacing 'D' made the bottom part zero, we used a special trick involving 'x' and differentiation to find the particular answer (P.I.). We then add these two parts for the complete solution.

🎯 Exam Tip: Always double-check if \( D \) substitution in the particular integral leads to a zero denominator, indicating the need for the \( x \cdot \frac{1}{f'(D)} \) rule.

Question 12. \( (3D^2 + D - 14) y = 13e^{2x} \)
Answer: Given the differential equation \( (3D^2 + D - 14) y = 13e^{2x} \).
First, find the auxiliary equation.
The auxiliary equation is \( 3m^2 + m - 14 = 0 \).
Factor this quadratic equation:
\( 3m^2 - 6m + 7m - 14 = 0 \)
\( 3m (m - 2) + 7 (m - 2) = 0 \)
\( (m - 2) (3m + 7) = 0 \)
This gives us two distinct roots:
\( m = 2 \)
\( 3m = -7 \implies m = -\frac{7}{3} \)
Since the roots are real and different, the complementary function (C.F) is \( Ae^{m_1 x} + Be^{m_2 x} \).
Therefore, the C.F. is \( Ae^{2x} + Be^{-\frac{7}{3}x} \).
Next, we find the particular integral (P.I) for \( 13e^{2x} \).
\( P.I = \frac{1}{3D^2 + D - 14} (13e^{2x}) \)
Substitute \( D = 2 \) (from the exponent). Check if the denominator is zero:
\( 3(2)^2 + (2) - 14 = 3(4) + 2 - 14 = 12 + 2 - 14 = 0 \).
Since the denominator is zero, we must use the special rule: multiply by \( x \) and differentiate the denominator with respect to \( D \).
\( P.I = x \frac{1}{6D + 1} (13e^{2x}) \)
Now, substitute \( D = 2 \) again into the new denominator:
\( 6(2) + 1 = 12 + 1 = 13 \). This is not zero.
So, \( P.I = x \frac{1}{13} (13e^{2x}) \)
\( P.I = x e^{2x} \).
The general solution is \( y = C.F + P.I \).
So, \( y = Ae^{2x} + Be^{-\frac{7}{3}x} + xe^{2x} \).
In simple words: We first find the basic solution (C.F) from the equation's left side. Then, for the right side, because a direct replacement didn't work, we used a special method with 'x' and differentiation to find the particular solution (P.I). We combine these parts for the complete answer.

🎯 Exam Tip: Be careful with the differentiation step when applying the special rule for particular integrals; an error here will lead to an incorrect P.I.

Question 13. Suppose that the quantity demanded \( Qd = 13 – 6p + 2\frac { dp }{dt} + \frac { d^2p }{dt^2} \) and quantity supplied \( Qs = -3 + 2p \) where \( p \) is the price. Find the equilibrium price for market clearence.
Answer: For market clearance, the quantity demanded must equal the quantity supplied.
So, we set \( Qd = Qs \).
\( 13 – 6p + 2\frac{dp}{dt} + \frac{d^2p}{dt^2} = -3 + 2p \)
Rearrange the terms to form a differential equation:
\( \frac{d^2p}{dt^2} + 2\frac{dp}{dt} - 6p - 2p + 13 + 3 = 0 \)
\( \frac{d^2p}{dt^2} + 2\frac{dp}{dt} - 8p + 16 = 0 \)
This is a non-homogeneous differential equation. We can write it as:
\( (D^2 + 2D - 8) p = -16 \), where \( D = \frac{d}{dt} \).
First, find the auxiliary equation for the complementary function.
The auxiliary equation is \( m^2 + 2m - 8 = 0 \).
Factor the quadratic equation:
\( (m + 4) (m - 2) = 0 \)
So, the roots are \( m = -4, 2 \). These are real and different roots.
The complementary function (C.F) is \( Ae^{m_1 t} + Be^{m_2 t} \). Here we use 't' as the independent variable.
Therefore, the C.F. is \( Ae^{-4t} + Be^{2t} \).
Next, we find the particular integral (P.I) for \( -16 \). We can write \( -16 \) as \( -16e^{0t} \).
\( P.I = \frac{1}{D^2 + 2D - 8} (-16e^{0t}) \)
Substitute \( D = 0 \). Check if the denominator is zero:
\( (0)^2 + 2(0) - 8 = -8 \). This is not zero.
So, \( P.I = \frac{-16}{-8} \)
\( P.I = 2 \).
The general solution for the price \( p \) is \( p = C.F + P.I \).
So, the equilibrium price function is \( p = Ae^{-4t} + Be^{2t} + 2 \). This equation describes how the price changes over time to reach equilibrium.
In simple words: For the market to be balanced, the amount people want to buy must be equal to the amount available. We set these two amounts equal, which gives us a special equation. Solving this equation helps us find a formula that describes how the price will move over time to reach a steady, balanced level.

🎯 Exam Tip: In market clearance problems, always set Quantity Demanded (Qd) equal to Quantity Supplied (Qs) to form the differential equation, and remember that constant terms on the right-hand side are treated as \( \text{constant} \cdot e^{0t} \) for P.I. calculations.