Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.4

Chapter 1

Question 1. Show that each of the following expressions is a solution of the corresponding given differential equation.
(i) \( y = 2x^2 \); \( xy' = 2y \)
(ii) \( y = ae^x + be^{-x} \); \( y'' - y = 0 \)
Answer:
(i) Given, \( y = 2x^2 \) ... (1)
First, we differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = 4x \)
This means \( y' = 4x \).
From this, we can find \( x = \frac{y'}{4} \).
Now, we substitute this value of \( x \) back into equation (1):
\( y = 2x(x) \)
\( y = 2x\left(\frac{y'}{4}\right) \)
\( y = \frac{xy'}{2} \)
By simplifying this equation, we get \( 2y = xy' \). This matches the given differential equation.
So, \( y = 2x^2 \) is a solution to the given differential equation \( xy' = 2y \). This shows how the original function satisfies the differential equation.
(ii) Given, \( y = ae^x + be^{-x} \) ... (1). The differential equation is \( y'' - y = 0 \).
First, we differentiate \( y \) with respect to \( x \):
\( y' = ae^x - be^{-x} \)
Next, we differentiate \( y' \) again with respect to \( x \) to find \( y'' \):
\( y'' = ae^x + be^{-x} \)
We notice that \( y'' \) is the same as the original \( y \). So, we can write \( y'' = y \). This is because differentiating \( e^x \) twice gives \( e^x \), and differentiating \( e^{-x} \) twice also gives \( e^{-x} \).
This means \( y'' - y = 0 \). This matches the given differential equation.
Therefore, \( y = ae^x + be^{-x} \) is a solution to the given differential equation \( y'' - y = 0 \).
In simple words: For each problem, we take the given function and find its derivatives (first and second, if needed). Then, we put these derivatives back into the differential equation given. If both sides of the differential equation become equal, it means the function is a solution.

🎯 Exam Tip: Remember to clearly show each differentiation step and substitute carefully to demonstrate that the given function satisfies the differential equation. Avoid skipping steps to ensure full marks.

Question 2. Find the value of m so that the function \( y = e^{mx} \) is a solution of the given differential equation.
(i) \( y' + 2y = 0 \)
(ii) \( y'' - 5y' + 6y = 0 \)
Answer:
(i) Given differential equation: \( y' + 2y = 0 \) ... (1)
Given function: \( y = e^{mx} \) ... (2)
First, differentiate equation (2) with respect to \( x \) to find \( y' \):
\( \frac{dy}{dx} = e^{mx} \cdot m \)
So, \( y' = me^{mx} \).
Now, substitute \( y' \) and \( y \) into the differential equation (1):
\( me^{mx} + 2(e^{mx}) = 0 \)
Factor out \( e^{mx} \):
\( e^{mx}(m + 2) = 0 \)
Since \( e^{mx} \) can never be zero, we must have:
\( m + 2 = 0 \)
\( m = -2 \)
Thus, the value of \( m \) is -2 for this function to be a solution. Finding this value helps us understand the specific properties required for the function to satisfy the equation.
(ii) Given differential equation: \( y'' - 5y' + 6y = 0 \) ... (1)
Given function: \( y = e^{mx} \) ... (2)
First, differentiate equation (2) with respect to \( x \) to find \( y' \):
\( \frac{dy}{dx} = e^{mx} \cdot m \)
So, \( y' = me^{mx} \).
Next, differentiate \( y' \) again with respect to \( x \) to find \( y'' \):
\( \frac{d^2y}{dx^2} = me^{mx} \cdot m \)
So, \( y'' = m^2e^{mx} \).
Now, substitute \( y'' \), \( y' \), and \( y \) into the differential equation (1):
\( m^2e^{mx} - 5(me^{mx}) + 6(e^{mx}) = 0 \)
Factor out \( e^{mx} \):
\( e^{mx}(m^2 - 5m + 6) = 0 \)
Since \( e^{mx} \) cannot be zero, we solve the quadratic equation:
\( m^2 - 5m + 6 = 0 \)
Factor the quadratic equation:
\( (m - 2)(m - 3) = 0 \)
This gives us two possible values for \( m \):
\( m = 2 \) or \( m = 3 \).
These values ensure that the function \( y = e^{mx} \) correctly solves the given differential equation. This type of equation often has multiple 'm' values.
In simple words: To find 'm', we first find the first and second derivatives of \( y = e^{mx} \). Then, we put these into the given differential equation. Since \( e^{mx} \) is never zero, we solve the remaining equation for 'm'. This tells us what 'm' needs to be for the function to work.

🎯 Exam Tip: When dealing with \( y = e^{mx} \) solutions, remember that \( e^{mx} \) can be factored out and then you solve the resulting auxiliary equation for \( m \). Be careful with the signs and coefficients during differentiation.

Question 3. The slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve.
Answer:
Let the slope of the tangent at any point \( (x, y) \) be \( \frac{dy}{dx} \).
The ordinate at that point is \( y \).
According to the problem, the slope is the reciprocal of four times the ordinate:
\( \frac{dy}{dx} = \frac{1}{4y} \)
To solve this differential equation, we separate the variables. We can rewrite the equation as:
\( 4y \cdot dy = dx \) ... (1)
Now, integrate both sides of equation (1):
\( \int 4y \cdot dy = \int dx \)
\( 4 \left(\frac{y^2}{2}\right) = x + C \)
\( 2y^2 = x + C \) ... (2)
We are given that the curve passes through the point \( (2, 5) \). We can use this to find the value of the constant \( C \).
Substitute \( x = 2 \) and \( y = 5 \) into equation (2):
\( 2(5)^2 = 2 + C \)
\( 2(25) = 2 + C \)
\( 50 = 2 + C \)
\( 50 - 2 = C \)
\( C = 48 \)
Finally, substitute the value of \( C \) back into equation (2) to get the equation of the curve:
\( 2y^2 = x + 48 \)
This equation uniquely describes the curve that fits the given conditions. Understanding the relationship between slope and coordinates is key here.
In simple words: The problem tells us how steep the curve is at any point. We write this as a math equation and then solve it by integrating. After that, we use the given point (2, 5) to find the missing number (C) in our equation, which gives us the final curve equation.

🎯 Exam Tip: For problems involving curve equations and slopes, always remember to set up the differential equation correctly, separate variables for integration, and use the given point to find the constant of integration.

Question 4. Show that \( y = e^{-x} + mx + n \) is a solution of the differential equation \( e^x \left(\frac{d^2y}{dx^2}\right) - 1 = 0 \).
Answer:
Given function: \( y = e^{-x} + mx + n \) ... (1)
Given differential equation: \( e^x \left(\frac{d^2y}{dx^2}\right) - 1 = 0 \)
First, differentiate equation (1) with respect to \( x \) to find \( y' \):
\( \frac{dy}{dx} = e^{-x}(-1) + m \)
\( y' = -e^{-x} + m \)
Next, differentiate \( y' \) again with respect to \( x \) to find \( y'' \):
\( \frac{d^2y}{dx^2} = -e^{-x}(-1) + 0 \)
\( \frac{d^2y}{dx^2} = e^{-x} \)
Now, substitute the value of \( \frac{d^2y}{dx^2} \) into the given differential equation:
\( e^x \left(e^{-x}\right) - 1 = 0 \)
When we multiply \( e^x \) by \( e^{-x} \), we add their exponents \( (x + (-x)) = 0 \), so \( e^0 = 1 \).
\( 1 - 1 = 0 \)
\( 0 = 0 \)
Since both sides are equal, it confirms that \( y = e^{-x} + mx + n \) is indeed a solution of the differential equation \( e^x \left(\frac{d^2y}{dx^2}\right) - 1 = 0 \). This demonstrates how certain functions fit specific differential equations perfectly.
In simple words: We take the given function, find its second derivative, and then plug that derivative into the differential equation. If the equation holds true (like 0 = 0), then the function is a correct solution.

🎯 Exam Tip: When showing a function is a solution, compute all required derivatives carefully. Remember that \( e^x \cdot e^{-x} = e^{x-x} = e^0 = 1 \).

Question 5. Show that \( y = ax + \frac{b}{x} \), where \( x \neq 0 \), is a solution of the differential equation \( x^2y'' + xy' - y = 0 \).
Answer:
Given function: \( y = ax + \frac{b}{x} \), where \( x \neq 0 \).
We can rewrite this as \( y = ax + bx^{-1} \).
Multiply by \( x \) to get \( xy = ax^2 + b \) ... (1)
First, differentiate equation (1) with respect to \( x \):
Using the product rule on the left side \( (xy)' = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y \).
\( x \frac{dy}{dx} + y = 2ax + 0 \)
So, \( xy' + y = 2ax \) ... (2)
Next, differentiate equation (2) again with respect to \( x \):
Using the product rule for \( xy' \), we get \( x \frac{d^2y}{dx^2} + y' \).
So, \( \left(x \frac{d^2y}{dx^2} + y'\right) + y' = 2a \)
\( x \frac{d^2y}{dx^2} + 2y' = 2a \) ... (3)
From equation (2), we know \( 2a = \frac{xy' + y}{x} \). Substitute this into equation (3):
\( x \frac{d^2y}{dx^2} + 2y' = \frac{xy' + y}{x} \)
Multiply the entire equation by \( x \) to clear the denominator:
\( x^2 \frac{d^2y}{dx^2} + 2xy' = xy' + y \)
Rearrange the terms to match the differential equation:
\( x^2 \frac{d^2y}{dx^2} + 2xy' - xy' - y = 0 \)
\( x^2 \frac{d^2y}{dx^2} + xy' - y = 0 \)
This matches the given differential equation. Therefore, \( y = ax + \frac{b}{x} \) is a solution to \( x^2y'' + xy' - y = 0 \). This step-by-step differentiation and substitution is key to verifying the solution.
In simple words: We start with the function, find its first and second derivatives. Then, we substitute these derivatives back into the given differential equation. If the equation simplifies to 0 = 0, it means the function is indeed a solution.

🎯 Exam Tip: When dealing with products like \( xy \) or \( xy' \), remember to apply the product rule carefully during differentiation to avoid errors.

Question 6. Show that \( y = ae^{-3x} + b \), where \( a \) and \( b \) are arbitrary constants, is a solution of the differential equation \( \frac{d^2y}{dx^2} + 3 \frac{dy}{dx} = 0 \).
Answer:
Given function: \( y = ae^{-3x} + b \) ... (1)
Given differential equation: \( \frac{d^2y}{dx^2} + 3 \frac{dy}{dx} = 0 \)
First, differentiate equation (1) with respect to \( x \) to find \( y' \):
\( \frac{dy}{dx} = ae^{-3x}(-3) + 0 \)
\( \frac{dy}{dx} = -3ae^{-3x} \) ... (2)
Next, differentiate equation (2) again with respect to \( x \) to find \( y'' \):
\( \frac{d^2y}{dx^2} = -3ae^{-3x}(-3) \)
\( \frac{d^2y}{dx^2} = 9ae^{-3x} \) ... (3)
Now, substitute the values of \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \) into the given differential equation:
\( 9ae^{-3x} + 3(-3ae^{-3x}) = 0 \)
\( 9ae^{-3x} - 9ae^{-3x} = 0 \)
\( 0 = 0 \)
Since the equation holds true, \( y = ae^{-3x} + b \) is a solution of the given differential equation \( \frac{d^2y}{dx^2} + 3 \frac{dy}{dx} = 0 \). The constants 'a' and 'b' represent arbitrary values that don't affect the differential equation's validity.
In simple words: We start with the given function and find its first and second derivatives. Then we put these into the differential equation. If the equation simplifies to zero, it means the function works as a solution.

🎯 Exam Tip: Pay close attention to chain rule application when differentiating exponential functions like \( e^{-3x} \), especially the negative sign and coefficients.

Question 7. Show that the differential equation representing the family of curves \( y^2 = 2a(x + a^{2/3}) \), where \( a \) is a positive parameter, is \( (y^2 - 2xy \frac{dy}{dx})^3 = 8(y \frac{dy}{dx})^5 \).
Answer:
Given family of curves: \( y^2 = 2a(x + a^{2/3}) \) ... (1)
First, differentiate equation (1) with respect to \( x \):
\( 2y \frac{dy}{dx} = 2a(1 + 0) \)
\( 2y \frac{dy}{dx} = 2a \)
From this, we can find the value of \( a \):
\( a = y \frac{dy}{dx} \) ... (2)
Now, substitute the value of \( a \) from equation (2) into the original equation (1):
\( y^2 = 2\left(y \frac{dy}{dx}\right) \left(x + \left(y \frac{dy}{dx}\right)^{2/3}\right) \)
\( y^2 = 2xy \frac{dy}{dx} + 2\left(y \frac{dy}{dx}\right)^{1 + 2/3} \)
\( y^2 = 2xy \frac{dy}{dx} + 2\left(y \frac{dy}{dx}\right)^{5/3} \)
Rearrange the terms to isolate the power term:
\( y^2 - 2xy \frac{dy}{dx} = 2\left(y \frac{dy}{dx}\right)^{5/3} \)
To remove the fractional exponent, we take the cubic power (raise to the power of 3) on both sides of the equation:
\( \left(y^2 - 2xy \frac{dy}{dx}\right)^3 = \left(2\left(y \frac{dy}{dx}\right)^{5/3}\right)^3 \)
\( \left(y^2 - 2xy \frac{dy}{dx}\right)^3 = 2^3 \left(\left(y \frac{dy}{dx}\right)^{5/3}\right)^3 \)
\( \left(y^2 - 2xy \frac{dy}{dx}\right)^3 = 8 \left(y \frac{dy}{dx}\right)^5 \)
This matches the given differential equation. Therefore, the differential equation represents the given family of curves. This method of eliminating the arbitrary constant is standard for forming differential equations.
In simple words: We start with the equation of the curves and differentiate it to get 'a'. Then we put 'a' back into the original curve equation. We simplify this new equation until it looks exactly like the differential equation given in the question. This shows they are the same.

🎯 Exam Tip: When forming a differential equation from a family of curves, the key is to eliminate the arbitrary constant(s) by differentiating and substituting. Be careful with fractional exponents and powers.

Question 8. Show that \( y = a \cos bx \) is a solution of the differential equation \( \frac{d^2y}{dx^2} + b^2y = 0 \).
Answer:
Given function: \( y = a \cos bx \) ... (1)
Given differential equation: \( \frac{d^2y}{dx^2} + b^2y = 0 \)
First, differentiate equation (1) with respect to \( x \) to find \( y' \):
\( \frac{dy}{dx} = a(-\sin bx)b \)
\( \frac{dy}{dx} = -ab \sin bx \) ... (2)
Next, differentiate equation (2) again with respect to \( x \) to find \( y'' \):
\( \frac{d^2y}{dx^2} = -ab(\cos bx)b \)
\( \frac{d^2y}{dx^2} = -ab^2 \cos bx \) ... (3)
From equation (1), we know that \( y = a \cos bx \). We can substitute this back into equation (3):
\( \frac{d^2y}{dx^2} = -b^2(a \cos bx) \)
So, \( \frac{d^2y}{dx^2} = -b^2y \).
Rearrange this to match the differential equation:
\( \frac{d^2y}{dx^2} + b^2y = 0 \)
This matches the given differential equation. Therefore, \( y = a \cos bx \) is a solution of \( \frac{d^2y}{dx^2} + b^2y = 0 \). This type of solution is common in problems involving simple harmonic motion.
In simple words: We start with the given function and find its first and second derivatives. Then, we put these derivatives into the differential equation. If the equation becomes true (like 0=0), then the function is proven to be a solution.

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating trigonometric functions like \( \cos bx \). The constant \( b \) will appear each time you differentiate.