Maharashtra Board Class 12 Maths Part 1 Chapter 8 Differential Equation 8.2 Solutions

Question 1.

Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) \( y = Ae^{3x} + Be^{-3x} \)
Answer:
Solution:
\( y = Ae^{3x} + Be^{-3x} \) ......(1)
Differentiating twice w.r.t. x, we get
\[ \frac{dy}{dx} = Ae^{3x} \times 3 + Be^{-3x} \times (-3) \]
\( \implies \frac{dy}{dx} = 3Ae^{3x} - 3Be^{-3x} \)
and
\[ \frac{d^2y}{dx^2} = 3Ae^{3x} \times 3 - 3Be^{-3x} \times (-3) \]
\( = 9Ae^{3x} + 9Be^{-3x} \)
\( = 9(Ae^{3x} + Be^{-3x}) = 9y \) .... [By (1)]
\( \implies \frac{d^2y}{dx^2} = 9y \)
This is the required D.E.
In simple words: We found the differential equation by differentiating the given equation twice with respect to x and then substituting the original equation back to eliminate the arbitrary constants A and B.

🎯 Exam Tip: Remember to differentiate until all arbitrary constants are eliminated, and then substitute back to form the final differential equation. Pay attention to constant elimination techniques.

Question 1.
(ii) \( y = C_2 + \frac{C_1}{X} \)
Answer:
Solution:
\( y = C_2 + \frac{C_1}{X} \)
.. \( xy = C_2X + C_1 \)
Differentiating w.r.t. x, we get
\[ x \frac{dy}{dx} + y \times 1 + 0 = C_2 \]
\( \implies x \frac{dy}{dx} + y = C_2 \)
Differentiating again w.r.t. x, we get
\[ x \left( \frac{d^2y}{dx^2} \right) + \frac{dy}{dx} \times 1 + \frac{dy}{dx} = 0 \]
\( \implies x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = 0 \)
is the required D.E.
In simple words: We eliminated the arbitrary constants by first multiplying by x to simplify, then differentiating twice with respect to x. The second differentiation removed the remaining constant.

🎯 Exam Tip: When constants are linear, algebraic manipulation followed by differentiation can simplify the process. Aim to eliminate constants one by one with each differentiation.

Question 1.
(iii) \( y = (C_1 + C_2x) e^x \)
Answer:
Solution:
\( y = (C_1 + C_2x) e^x \)
.. \( e^{-x} y = C_1 + C_2X \)
Differentiating w.r.t. x, we get
\[ e^{-x} \frac{dy}{dx} + y e^{-x} (-1) = 0 + C_2 \times 1 \]
\( \implies e^{-x} \left( \frac{dy}{dx} - y \right) = C_2 \)
Differentiating again w.r.t. x, we get
\[ e^{-x} \left( \frac{d^2y}{dx^2} \right) + \left( \frac{dy}{dx} - y \right) e^{-x} (-1) = 0 \]
\( \implies e^{-x} \left( \frac{d^2y}{dx^2} - \frac{dy}{dx} + y \right) = 0 \)

\( \implies \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + y = 0 \)
This is the required D.E.
In simple words: The equation was simplified by multiplying by \( e^{-x} \) and then differentiated twice. Each differentiation eliminated a constant, leading to the final second-order differential equation.

🎯 Exam Tip: For equations involving products with exponential terms, isolating the arbitrary constants before differentiation can simplify the process. Remember to apply the product rule carefully.

Question 1.
(iv) \( y = c_1 e^{3x} + c_2 e^{2x} \)
Answer:
Solution:
\( y = c_1 e^{3x} + c_2 e^{2x} \) .... (1)
Differentiating twice w.r.t. x, we get
\[ \frac{dy}{dx} = 3c_1 e^{3x} + 2c_2 e^{2x} \] .... (2)
\[ \frac{d^2y}{dx^2} = 9c_1 e^{3x} + 4c_2 e^{2x} \] .... (3)
These three equations in \( c_1e^{3x} \) and \( c_2e^{2x} \) are consistent.

\( \implies \) determinant of their consistency condition is zero.

\( \implies y(12-18) - \left( \frac{dy}{dx} \right)(4-2) + 1(9-3) = 0 \)

\( \implies -6y - 4 \frac{dy}{dx} + 2 \frac{dy}{dx} + 9 \frac{dy}{dx} - 3 \frac{d^2y}{dx^2} = 0 \)

\( \implies \frac{d^2y}{dx^2} - 5 \frac{dy}{dx} + 6y = 0 \)
This is the required D.E.
Alternative Method:
\( y = c_1e^{3x} + c_2e^{2x} \)
Dividing both sides by \( e^{2x} \), we get
\( e^{-2x}y = C_1e^x + C_2 \)
Differentiating w.r.t. x, we get
\[ e^{-2x} \frac{dy}{dx} + y e^{-2x} (-2) = c_1e^x + 0 \]
\( \implies e^{-2x} \left( \frac{dy}{dx} - 2y \right) = c_1e^x \)
Dividing both sides by \( e^x \), we get
\( e^{-3x} \left( \frac{dy}{dx} - 2y \right) = c_1 \)
Differentiating w.r.t. x, we get
\[ e^{-3x} \left( \frac{d^2y}{dx^2} \right) + \left( \frac{dy}{dx} - 2y \right) e^{-3x} (-3) = 0 \]
\( \implies e^{-3x} \left( \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} - 3 \frac{dy}{dx} + 6y \right) = 0 \)

\( \implies \frac{d^2y}{dx^2} - 5 \frac{dy}{dx} + 6y = 0 \)
This is the required D.E.
In simple words: This alternative approach involves strategically manipulating the equation by dividing by exponential terms to simplify and isolate constants before differentiation. Each differentiation then helps eliminate a constant until the final differential equation is obtained.

🎯 Exam Tip: When dealing with multiple exponential constants, an alternative method involving division by exponential terms can sometimes lead to a quicker solution. Be methodical with algebraic rearrangement and differentiation steps.

Question 1.
(v) \( y^2 = (x + c)^3 \)
Answer:
Solution:
\( y^2 = (x + c)^3 \)
Differentiating w.r.t. x, we get
\[ 2y \frac{dy}{dx} = 3(x + c)^2 \cdot (1) = 3 (x + c)^2 \]
\( \implies (x + c)^2 = \frac{2y}{3} \frac{dy}{dx} \)

\( \implies (x + c) = \left( \frac{2y}{3} \frac{dy}{dx} \right)^{1/2} \)

\( \implies (y^2)^2 = \frac{8y^3}{27} \left( \frac{dy}{dx} \right)^3 \) ...... [By (1)]

\( \implies 27y^4 = 8y^3 \left( \frac{dy}{dx} \right)^3 \)

\( \implies 27y = 8 \left( \frac{dy}{dx} \right)^3 \)

\( \implies \left( \frac{dy}{dx} \right)^3 = \frac{27y}{8} \)

\( \implies \frac{dy}{dx} = \left( \frac{27y}{8} \right)^{1/3} \)
This is the required D.E.
In simple words: We eliminated the constant by differentiating the equation once, then expressing \((x+c)\) in terms of \((y)\) and \((dy/dx)\), and finally substituting back into the original equation to remove \((x+c)\) entirely.

🎯 Exam Tip: For implicit equations with constants, careful use of substitution after differentiation is key. Simplify the expressions involving the constant before substituting to avoid algebraic complexity.

Question 2.
Find the differential equation by eliminating arbitrary constant from the relation \( x^2 + y^2 = 2ax \).
Answer:
Solution:
\( x^2 + y^2 = 2ax \) ....(1)
Differentiating both sides w.r.t. x, we get
\[ 2x + 2y \frac{dy}{dx} = 2a \] Substituting value of \( 2a \) in equation (1), we get
\( x^2 + y^2 = \left[ 2x + 2y \frac{dy}{dx} \right] x = 2x^2 + 2xy \frac{dy}{dx} \)

\( \implies 2xy \frac{dy}{dx} = y^2 - x^2 \)
is the required D.E.
In simple words: We eliminated the constant by differentiating the given equation with respect to x, then substituting the expression for the constant back into the original equation to form the differential equation.

🎯 Exam Tip: When the constant is linear (e.g., `2ax`), differentiating and directly substituting the constant's value back into the original equation is often the most straightforward method. Label equations to keep track of substitutions.

Question 3.
Form the differential equation by eliminating arbitrary constants from the relation \( bx + ay = ab \).
Answer:
Solution:
\( bx + ay = ab \)

\( \implies ay = -bx + ab \)

\( \implies y = -\frac{b}{a} x + b \)
Differentiating w.r.t. x, we get
\[ \frac{dy}{dx} = -\frac{b}{a} \times 1 + 0 = -\frac{b}{a} \] Differentiating again w.r.t. x, we get
\[ \frac{d^2y}{dx^2} = 0 \] is the required D.E.
In simple words: The equation was first rearranged to express y, then differentiated twice. Since the constants are linear, a second differentiation immediately resulted in a zero second derivative, thus eliminating the constants.

🎯 Exam Tip: For linear equations involving arbitrary constants, one or two differentiations are usually sufficient to eliminate them. The order of the differential equation will often be equal to the number of independent arbitrary constants.

Question 4.
Find the differential equation whose general solution is \( x^3 + y^3 = 35ax \).
Answer:
Solution:
\( x^3 + y^3 = 35ax \) ...(i)
Differentiating w.r.t. x, we get
\[ 3x^2 + 3y^2 \frac{dy}{dx} = 35a \] ...(ii)
Substituting (ii) in (i), we get
\( x^3 + y^3 = \left( 3x^2 + 3y^2 \frac{dy}{dx} \right) x \)

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

\( \implies 2x^3 - y^3 + 3xy^2 \frac{dy}{dx} = 0 \), which is the required differential equation.
In simple words: We found the differential equation by differentiating the given general solution once to find an expression for the arbitrary constant, and then substituted this back into the original equation to eliminate the constant.

🎯 Exam Tip: When the arbitrary constant is clearly expressed after differentiation, substituting it back into the original equation is a common strategy. Be careful with algebraic rearrangement to simplify the final differential equation.

Question 5.
Form the differential equation from the relation \( x^2 + 4y^2 = 4b^2 \).
Answer:
Solution:
\( x^2 + 4y^2 = 4b^2 \)
Differentiating w.r.t. x, we get
\[ 2x + 4 \left( 2y \frac{dy}{dx} \right) = 0 \]
\( \implies x + 4y \frac{dy}{dx} = 0 \)
is the required D.E.
In simple words: We formed the differential equation by differentiating the given relation once with respect to x. Since there was only one arbitrary constant and it disappeared after the first differentiation, the resulting equation is the required differential equation.

🎯 Exam Tip: If the arbitrary constant vanishes after a single differentiation, that first derivative directly gives the required first-order differential equation. Always simplify the resulting equation.