ML Aggarwal Class 12 Maths Solutions Section A Chapter 15 Differential Equations

15. Differential Equations

Introduction

In fields like engineering, physics, chemistry, and sometimes economics and biology, it becomes necessary to build a mathematical framework to represent specific problems. Often, these mathematical frameworks require finding an unknown function (or functions) that meets an equation holding the derivatives of that unknown function (or functions). Such equations are known as differential equations. The goal of this chapter is to work out differential equations - that is, to find the unknown function (or functions) that fulfills the given differential equation.

15.1 Differential Equation

An equation containing unknown functions and their derivatives with respect to one or more independent variables is called a differential equation.

An equation that contains an unknown function or functions, holding only one independent variable and derivatives relative to that independent variable is called an ordinary differential equation.

In this book, we will work only with ordinary differential equations.

15.1.1 Order and Degree of a Differential Equation

The order of a differential equation is the order of the highest order derivative of the dependent variable relative to the independent variable appearing in the equation. When each term holding derivatives of a differential equation is a polynomial (or can be restated as polynomial), then the exponent of the highest order derivative is called the degree of the differential equation.

When any term of a differential equation cannot be stated as a polynomial in the derivative (or derivatives), then the degree of the differential equation is not defined.

For example, consider the equations:

(i) \( \frac{dy}{dx} = 2x^3 + 7\sqrt{x} \)
(ii) \( 3\left(\frac{dy}{dx}\right)^2 = \sin 2x - \frac{2}{\frac{dy}{dx}} \)
(iii) \( 5\frac{d^2y}{dx^2} + 8\left(\frac{dy}{dx}\right)^2 = \log x \)
(iv) \( y = x\frac{dy}{dx} + a\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \)
(v) \( \left[1 - \left(\frac{dy}{dx}\right)^2\right]^{3/2} = k\frac{d^2y}{dx^2} \)
(vi) \( (3x^2 + 5y)dy - 5x \, dx = 0 \)
(vii) \( (y')^2 + \cos^2 y = 0 \)
(viii) \( (y'')^2 - \sin y' = 0 \)
(ix) \( y''' - 3(y'')^2 + (y')^3 + \log y = 5x \)

These are all ordinary differential equations.

Analysis of Each Equation:

(i) The maximum order derivative present is \( \frac{dy}{dx} \), so the order is 1. The term in the derivative forms a polynomial, so its degree is the highest exponent of \( \frac{dy}{dx} \), which is 1. Thus, the degree is 1.

(ii) The maximum order derivative present is \( \frac{dy}{dx} \), so the order is 1. It can be rewritten as \( 3\left(\frac{dy}{dx}\right)^3 = \sin 2x \cdot \frac{dy}{dx} - 2 \). Since each term in the derivatives forms a polynomial, its degree is the highest exponent of \( \frac{dy}{dx} \), which is 3. Thus, the degree is 3.

(iii) The maximum order derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2. Since each term in the derivatives forms a polynomial, its degree is the highest exponent of \( \frac{d^2y}{dx^2} \), which is 1. Thus, the degree is 1.

(iv) The maximum order derivative present is \( \frac{dy}{dx} \), so the order is 1. It can be written as \( \left(y - x\frac{dy}{dx}\right)^2 = a^2\left[1 + \left(\frac{dy}{dx}\right)^2\right] \) or \( (x^2 - a^2)\left(\frac{dy}{dx}\right)^2 - 2xy\frac{dy}{dx} + y^2 - a^2 = 0 \). Since each term in the derivatives forms a polynomial, its degree is the highest exponent of \( \frac{dy}{dx} \), which is 2. Thus, the degree is 2.

(v) The maximum order derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2. It can be written as \( \left[1 - \left(\frac{dy}{dx}\right)^2\right]^3 = k^2\left(\frac{d^2y}{dx^2}\right)^2 \). Since each term in the derivative forms a polynomial, its degree is the highest exponent of \( \frac{d^2y}{dx^2} \), which is 2. Thus, the degree is 2.

(vi) The given differential equation can be written as \( (3x^2 + 5y)\frac{dy}{dx} = 5x \), so this equation has order 1 and degree 1.

(vii) The maximum order derivative present is \( y' \), so the order is 1. The term holding the derivative forms a polynomial, so its degree is the highest exponent of \( y' \), which is 2. Thus, the degree is 2. Note that the term \( \cos^2 y \) is not a polynomial in \( y \).

(viii) The maximum order derivative present is \( y'' \), so the order is 2. The term \( \sin y' \) is not a polynomial in \( y' \), so the degree of the given differential equation is not defined.

(ix) The maximum order derivative present is \( y''' \), so the order is 3. Each term holding derivatives \( y''' \), \( y'' \), and \( y' \) are polynomials, so its degree is the highest exponent of \( y''' \), which is 1. Thus, the degree is 1. Note that the term \( \log y \) is not a polynomial in \( y \).

In general, an ordinary differential equation of order one has the form \( F\left(x, y, \frac{dy}{dx}\right) = 0 \) where F is a function of the variables \( x \), \( y \), and \( \frac{dy}{dx} \); and an ordinary differential equation of order 2 has the form \( F\left(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}\right) = 0 \), etc.

More generally, an equation of the form \( F(x, y, y_1, y_2, \ldots, y_n) = 0 \) is called an ordinary differential equation of order n.

15.1.2 Linear Differential Equation

An ordinary differential equation \( F(x, y, y_1, y_2, \ldots, y_n) = 0 \) is called linear if and only if the function F is a linear function of the variables \( y, y_1, y_2, \ldots, y_n \) - that is, if and only if the dependent variable and its derivatives appear only in the first degree and are not multiplied with each other. Otherwise, it is called non-linear.

Therefore, a general linear differential equation of order n may be expressed as

\( P_0\frac{d^ny}{dx^n} + P_1\frac{d^{n-1}y}{dx^{n-1}} + P_2\frac{d^{n-2}y}{dx^{n-2}} + \ldots + P_ny = Q \) ...(i)

where \( P_0, P_1, P_2, \ldots, P_n, Q \) are all functions of \( x \) only and \( P_0 \neq 0 \).

The linear differential equation (i) is called homogeneous if Q is a zero function and it is called non-homogeneous if Q is a non-zero function.

Note that a linear differential equation is always of degree one but every differential equation of degree one need not be linear. For example, the differential equation \( x\frac{d^2y}{dx^2} - 5\sin\left(\frac{dy}{dx}\right) = 3y \) has degree 1 but it is not linear.

Exercise 15.1

Determine the order and the degree (if any) of each of the following differential equations. Also state if these are linear or non-linear.

Question 1. \( \frac{dy}{dx} + 5y = 0 \)
Answer: Order 1, Degree 1; linear.
In simple words: The highest derivative is \( \frac{dy}{dx} \) (order 1), raised to power 1 (degree 1), and the dependent variable \( y \) appears to the first power only - so it is linear.

Exam Tip: For simple first-order equations without products or powers, always check that the dependent variable and its derivative each appear only to the first degree.

Question 2. \( \frac{dy}{dx} + 2y = 7x^3 \)
Answer: Order 1, Degree 1; linear.
In simple words: The highest derivative is first order, the exponent is 1, and \( y \) appears to the first power - this is a linear equation.

Exam Tip: The right-hand side being a polynomial in \( x \) does not affect linearity - only the status of \( y \) and its derivatives matters.

Question 3. \( (x^2y - 3x)dy + (x^3 - 3y^2)dx = 0 \)
Answer: Order 1, Degree 1; non-linear.
In simple words: Rearranging shows the first derivative to the power 1, but products like \( xy \) and powers like \( y^2 \) appear - making it non-linear.

Exam Tip: When a dependent variable is multiplied by independent variables or appears with exponents greater than one, the equation is non-linear.

Question 4. \( \sqrt{1 - y^2} \, dx + \sqrt{1 - x^2} \, dy = 0 \)
Answer: Order 1, Degree 1; non-linear.
In simple words: The terms \( \sqrt{1 - y^2} \) and \( \sqrt{1 - x^2} \) mean \( y \) appears inside a radical - this creates a non-polynomial form, making it non-linear.

Exam Tip: Any derivative or variable appearing under a radical, inside a trigonometric function, or in a logarithm signals non-linearity.

Question 5. \( \frac{1}{x}\frac{d^2y}{dx^2} + 5x\frac{dy}{dx} = \sin 2x \)
Answer: Order 2, Degree 1; linear.
In simple words: The highest derivative is second order, each derivative appears to the first power, and \( y \) appears only to the first power - so it is linear.

Exam Tip: Coefficients that are functions of \( x \) alone do not prevent linearity - check only that derivatives and the dependent variable are to the first power.

Question 6. \( y' + 6y^2 + y = 0 \)
Answer: Order 1, Degree 1; non-linear.
In simple words: The term \( 6y^2 \) means \( y \) appears to the second power - this violates the first-degree requirement, making the equation non-linear.

Exam Tip: The presence of any dependent variable raised to a power higher than one immediately signals non-linearity, regardless of other terms.

Question 7. \( (y')^2 + y^2 - 1 = 0 \)
Answer: Order 1, Degree 2; non-linear.
In simple words: The highest derivative is first order (degree is the highest exponent of that derivative), but \( (y')^2 \) means it is raised to power 2 - so the degree is 2. The \( y^2 \) term also makes it non-linear.

Exam Tip: Degree refers to the exponent of the highest derivative, not the dependent variable - here the derivative is squared, so degree is 2.

Question 8. \( (y')^2 + y^3 + y = 0 \)
Answer: Order 1, Degree 2; non-linear.
In simple words: The derivative appears squared (degree 2), and the dependent variable appears cubed and to the first power - both features make it non-linear.

Exam Tip: Non-linearity can arise from either powers of derivatives or powers of the dependent variable - both must be checked.

Question 9. \( y'' + y^2 = 0 \)
Answer: Order 2, Degree 1; non-linear.
In simple words: The highest derivative is second order (order 2), each derivative is to the first power (degree 1), but \( y^2 \) makes the equation non-linear.

Exam Tip: Linearity is lost whenever the dependent variable or any of its derivatives is raised to a power other than one or multiplied together.

Question 10. \( \left(\frac{d^2x}{dt^2}\right)^2 - 7\left(\frac{dx}{dt}\right)^3 = \log t \)
Answer: Order 2, Degree 2; non-linear.
In simple words: The highest derivative is second order, and that second derivative is raised to the power 2, giving degree 2. The term \( \left(\frac{dx}{dt}\right)^3 \) also makes it non-linear.

Exam Tip: When the highest derivative is squared or raised to any power greater than one, that power becomes the degree of the equation.

Question 11. \( 3x\frac{dy}{dx} + \frac{5}{\frac{dy}{dx}} = y^3 \)
Answer: Order 1, Degree 2; non-linear.
In simple words: The term \( \frac{5}{\frac{dy}{dx}} \) can be rewritten as \( 5\left(\frac{dy}{dx}\right)^{-1} \), so the degree is 2 when we clear the denominator and express as a polynomial. The \( y^3 \) term adds non-linearity.

Exam Tip: When derivatives appear in a denominator, rewrite the equation so derivatives are in the numerator before determining degree.

Question 12. \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = 2x - \frac{dy}{dx} \)
Answer: Order 1, Degree 1; linear.
In simple words: After squaring both sides to remove the radical, the equation can be written with the derivative to the first power and the dependent variable to the first power, making it linear.

Exam Tip: Always try to rewrite the equation in polynomial form by clearing radicals before assigning degree.

Question 13. \( y' + \cos y' = 0 \)
Answer: Order 1, Degree not defined; non-linear.
In simple words: The term \( \cos y' \) is not a polynomial in the derivative - you cannot express this as a polynomial no matter how you rearrange - so the degree is not defined.

Exam Tip: Whenever a derivative appears inside a trigonometric function, logarithm, or exponential, the degree is not defined.

Question 14. \( y'' + 2y' + \sin y = 0 \)
Answer: Order 2, Degree 1; non-linear.
In simple words: The highest derivative is second order (order 2) and is to the first power (degree 1), but \( \sin y \) is not a polynomial in \( y \) - making it non-linear.

Exam Tip: A dependent variable inside any trigonometric function makes the equation non-linear, even if the derivatives themselves are to the first power.

Question 15. \( \frac{d^3y}{dx^3} + 7x^2\frac{dy}{dx} = 5x^2 - 2y \)
Answer: Order 3, Degree 1; linear.
In simple words: The highest derivative is third order (order 3), all derivatives and the dependent variable are to the first power (degree 1), so it is a linear equation.

Exam Tip: Higher-order derivatives can still be part of a linear equation as long as they and the dependent variable are to the first power only and not multiplied together.

Question 16. \( \frac{d^2y}{dx^2} = \sin x + \cos x \)
Answer: Order 2, Degree 1; linear.
In simple words: The highest derivative is second order, it is to the first power, and the dependent variable does not appear - so it is a linear equation.

Exam Tip: An equation with only derivatives (no dependent variable term) is still linear as long as each derivative is to the first power.

Question 17. \( \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = 5\frac{d^2y}{dx^2} \)
Answer: Order 2, Degree 2; non-linear.
In simple words: The left side contains a power (exponent 3/2) applied to a term with \( \frac{dy}{dx} \), which when expanded or cleared gives the second derivative to the power 2 - hence degree 2 and non-linear.

Exam Tip: Terms with roots or fractional exponents applied to derivatives or dependent variables typically signal non-linearity after the equation is simplified.

Question 18. \( (y''')^2 + (y'')^3 + (y')^4 + y^5 = 0 \)
Answer: Order 3, Degree 2; non-linear.
In simple words: The highest derivative is third order (order 3). That derivative is squared (degree 2 based on the highest derivative's exponent). The other terms with higher powers also make it non-linear.

Exam Tip: The degree is determined by the exponent of the highest-order derivative only, not the exponents of lower-order derivatives or the dependent variable.

Question 19. \( y^{(iv)} + y''' + y'' + y' + y = 0 \)
Answer: Order 4, Degree 1; linear.
In simple words: The highest derivative is fourth order (order 4), every derivative and the dependent variable is to the first power (degree 1), and they are added - making it linear.

Exam Tip: As long as all derivatives are to the first power and are added (not multiplied together), the equation remains linear no matter how many derivatives are present.

Question 20. \( y^{(v)} + y^2 + e^{y'} = 0 \)
Answer: Order 5, Degree not defined; non-linear.
In simple words: The highest derivative is fifth order (order 5). However, \( y' \) appears in the exponent of e, which cannot be written as a polynomial - so the degree is not defined.

Exam Tip: Exponential and logarithmic functions of derivatives or the dependent variable always lead to an undefined degree.

Question 21. \( y = px + a\sqrt{p^2 + b^2} \) where \( p = \frac{dy}{dx} \)
Answer: Order 1, Degree 2; non-linear.
In simple words: Substituting \( p = \frac{dy}{dx} \) and clearing the radical by squaring produces an equation where the derivative is squared (degree 2). The radical form and the multiplication of terms make it non-linear.

Exam Tip: For equations given in parametric form with a parameter \( p \) representing the derivative, always substitute and simplify to identify order and degree.

15.2 Solution of a Differential Equation

A solution of a differential equation is a relationship between the variables, through which and the derivatives obtained from it, the equation is satisfied.

For example:

(i) Consider the differential equation \( \frac{dy}{dx} = \frac{1}{x} \), \( x > 0 \). Then \( y = \log x \) is a solution of this equation, and a more general solution of this differential equation is \( y = A + \log x \), where A is an arbitrary constant.

(ii) Consider the differential equation \( \frac{d^2y}{dx^2} + y = 0 \) ...(1)

Let \( y = A \sin x + B \cos x \) ...(2), where A and B are arbitrary constants.

Differentiating (2) twice relative to x, we obtain

\( \frac{dy}{dx} = A \cos x - B \sin x \) and \( \frac{d^2y}{dx^2} = -A \sin x - B \cos x \)

\( \Rightarrow \frac{d^2y}{dx^2} = -y \) [using (2)] \( \Rightarrow \frac{d^2y}{dx^2} + y = 0 \)

Therefore, \( y = A \sin x + B \cos x \) is a solution of the differential equation (1).

It can be easily verified that \( y = 2 \sin x - 3 \cos x \), \( y = A \sin x \), and \( y = B \cos x \) are all solutions of (1). The most general of these solutions is (2); all others are particular solutions of the differential equation (1).

General Solution

Let the equation holding the variables x, y and n independent arbitrary constants be

\( f(x, y, c_1, c_2, \ldots, c_n) = 0 \) ...(i)

and the differential equation obtained from (i) be

\( F\left(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots, \frac{d^ny}{dx^n}\right) = 0 \) ...(ii)

then (i) is called the general solution (or complete primitive or complete solution) of (ii).

Note that the general solution of an ordinary differential equation of nth order holds n independent arbitrary constants. Therefore, the general solution of an ordinary differential equation of order one holds one arbitrary constant and of second order holds two independent arbitrary constants and so on.

Particular Solution

Any solution obtained from the general solution of a differential equation by assigning particular values to some or all the arbitrary constants is called a particular solution (or particular primitive).

The problem of finding the particular solution of a differential equation that fulfills the given condition(s) is called an initial value problem.

Illustrative Examples

Question. Show that \( y = \frac{A}{x + A} \) is a solution of the differential equation \( xy_1 + y = y^2 \).
Answer: Given \( y = \frac{A}{x + A} \Rightarrow y(x + A) = A \) ...(i)

Differentiating relative to x, we get

\( y \cdot 1 + (x + A)y_1 = 0 \Rightarrow x + A = -\frac{y}{y_1} \)

\( \Rightarrow A = -x - \frac{y}{y_1} \)

Substituting the values of \( x + A \) and \( A \) in (i), we get

\( y\left(-\frac{y}{y_1}\right) = -x - \frac{y}{y_1} \Rightarrow y^2 = xy_1 + y \)

Hence, \( y = \frac{A}{x + A} \) is a solution of the given differential equation.
In simple words: We substitute the given function into the differential equation and verify that it satisfies the equation by showing both sides are equal.

Exam Tip: To prove a function is a solution, differentiate it to the required order, then substitute into the original equation and confirm both sides match.

Question. Verify that \( y = A \cos x - B \sin x \) is a solution of the differential equation \( \frac{d^2y}{dx^2} + y = 0 \).
Answer: Given \( y = A \cos x - B \sin x \) ...(i)

Differentiating (i) twice relative to x, we get

\( \frac{dy}{dx} = -A \sin x - B \cos x \) and \( \frac{d^2y}{dx^2} = -A \cos x - B(-\sin x) = -(A \cos x - B \sin x) = -y \) (using (i))

\( \Rightarrow \frac{d^2y}{dx^2} + y = 0 \)

Therefore, \( y = A \cos x - B \sin x \) is a solution of the given differential equation.
In simple words: We differentiate the given function twice, substitute into the equation, and confirm that the left side simplifies to zero, verifying it is a solution.

Exam Tip: For second-order equations, always differentiate twice and substitute both derivatives before concluding the verification.

Question. Show that \( y = Ax + \frac{B}{x} \) is a solution of the differential equation \( x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = 0 \).
Answer: Given \( y = Ax + \frac{B}{x} \Rightarrow xy = Ax^2 + B \)

Differentiating relative to x, we get

\( x\frac{dy}{dx} + y = 2Ax \Rightarrow \frac{dy}{dx} + \frac{y}{x} = 2A \)

Differentiating relative to x, we get

\( \frac{d^2y}{dx^2} + \frac{x\frac{dy}{dx} - y \cdot 1}{x^2} = 0 \Rightarrow x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = 0 \)

Hence, \( y = Ax + \frac{B}{x} \) is a solution of \( x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = 0 \).
In simple words: We differentiate the function step by step, then substitute the derivatives into the equation to confirm it equals zero.

Exam Tip: When derivatives involve quotients or products, use the product and quotient rules carefully before substituting into the equation.

Question. Show that \( y = ae^{2x} + be^{-x} \) is a solution of the differential equation \( \frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0 \).
Answer: Given \( y = ae^{2x} + be^{-x} \) ...(i)

Differentiating (i) relative to x, we get \( \frac{dy}{dx} = ae^{2x} \cdot 2 + be^{-x}(-1) \) ...(ii)

Adding (i) and (ii), we get \( \frac{dy}{dx} + y = 3ae^{2x} \) ...(iii)

Differentiating (iii) relative to x, we get \( \frac{d^2y}{dx^2} + \frac{dy}{dx} = 3a \cdot e^{2x} \cdot 2 \) ...(iv)

Multiplying (iii) by 2 and subtracting from (iv), we get \( \frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0 \)

Hence, \( y = ae^{2x} + be^{-x} \) is a solution of the given differential equation.
In simple words: We take derivatives of the exponential function, combine them using addition and subtraction of equations to eliminate constants, then verify the result matches the original equation.

Exam Tip: For exponential solutions, often combining equations algebraically (adding or multiplying) helps eliminate specific terms and verify the solution.

Question. Show that \( y = (a + bx)e^{2x} \) is a solution of the differential equation \( y_2 - 4y_1 + 4y = 0 \).
Answer: Given \( y = (a + bx)e^{2x} \) ...(i)

Differentiating (i) relative to x, we get

\( y_1 = (a + bx)e^{2x} \cdot 2 + e^{2x}(0 + b \cdot 1) \Rightarrow y_1 = 2y + be^{2x} \) ...(ii) (using (i))

Differentiating again relative to x, we get

\( y_2 = 2y_1 + be^{2x} \cdot 2 \)

\( \Rightarrow y_2 = 2y_1 + 2(y_1 - 2y) \) (using (ii))

\( \Rightarrow y_2 - 4y_1 + 4y = 0 \)

Hence, \( y = (a + bx)e^{2x} \) is a solution of the given differential equation.
In simple words: We differentiate the product using the product rule twice, substitute into the equation, and verify that all terms combine to give zero.

Exam Tip: For products involving polynomials and exponentials, use the product rule consistently and simplify by substituting earlier derivatives into later ones.

Question. Show that \( y = e^{m\cos^{-1}x} \) is a solution of the differential equation \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - m^2y = 0 \).
Answer: Given \( y = e^{m\cos^{-1}x} \) ...(i)

Differentiating relative to x, we get

\( \frac{dy}{dx} = e^{m\cos^{-1}x} \cdot m \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{m}{\sqrt{1-x^2}} \cdot y \) (using (i))

\( \Rightarrow \sqrt{1-x^2} \frac{dy}{dx} = -my \Rightarrow (1-x^2)\left(\frac{dy}{dx}\right)^2 = m^2y^2 \)

Differentiating again relative to x, we get

\( (1-x^2) \cdot 2\frac{dy}{dx} \cdot \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \cdot (-2x) = m^2 \cdot 2y \cdot \frac{dy}{dx} \)

Dividing both sides by \( 2\frac{dy}{dx} \), we get

\( (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - m^2y = 0 \)

Hence, \( y = e^{m\cos^{-1}x} \) is a solution of \( (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - m^2y = 0 \).
In simple words: We differentiate the composite function using the chain rule, square the first derivative to eliminate the square root, then differentiate again and simplify to obtain the given equation.

Exam Tip: For inverse trigonometric functions in the exponent, remember their derivatives (e.g., \( \frac{d}{dx}[\cos^{-1}x] = -\frac{1}{\sqrt{1-x^2}} \)) and use the chain rule carefully.

Exercise 15.2

In each of the following (1 to 4), show that the given function is a solution of the differential equation:

Question 1. \( \frac{dy}{dx} + y = 0 \); \( y = Ae^{-x} \)
Answer: Given \( y = Ae^{-x} \)

Differentiating, \( \frac{dy}{dx} = -Ae^{-x} = -y \)

Therefore, \( \frac{dy}{dx} + y = -y + y = 0 \), confirming that \( y = Ae^{-x} \) satisfies the equation.
In simple words: When we differentiate the exponential function, we get the negative of itself, so adding the original function to its derivative gives zero.

Exam Tip: For exponential functions, differentiation is straightforward - just bring down the exponent's derivative as a coefficient.

Question 2. \( y_2 + 9y = 0 \); \( y = 4\sin 3x \)
Answer: Given \( y = 4\sin 3x \)

Differentiating twice: \( y_1 = 4 \cdot 3\cos 3x = 12\cos 3x \) and \( y_2 = 12 \cdot (-3)\sin 3x = -36\sin 3x = -9 \cdot 4\sin 3x = -9y \)

Therefore, \( y_2 + 9y = -9y + 9y = 0 \), confirming the solution.
In simple words: When differentiating trigonometric functions twice with a coefficient, the function returns with a negative sign and a multiplier equal to the square of the coefficient.

Exam Tip: For \( \sin(kx) \) or \( \cos(kx) \), differentiating twice produces the original function multiplied by \( -k^2 \).

Question 3. \( y_2 + 4y = 0 \); \( y = A\cos 2x - B\sin 2x \)
Answer: Given \( y = A\cos 2x - B\sin 2x \)

Differentiating: \( y_1 = -2A\sin 2x - 2B\cos 2x \)

Differentiating again: \( y_2 = -4A\cos 2x + 4B\sin 2x = -4(A\cos 2x - B\sin 2x) = -4y \)

Therefore, \( y_2 + 4y = -4y + 4y = 0 \), confirming the solution.
In simple words: A linear combination of sine and cosine functions (with matching coefficients inside) produces a second derivative that is the negative of four times the original function.

Exam Tip: When a solution is a combination of \( \sin(kx) \) and \( \cos(kx) \), the second derivative automatically gives \( -k^2 \) times the original function.

Question 4. \( y = x\frac{dy}{dx} + a\frac{dy}{dx} \); \( y = cx + \frac{a}{c} \)
Answer: Given \( y = cx + \frac{a}{c} \)

Differentiating: \( \frac{dy}{dx} = c \)

Substituting into the equation: \( cx + \frac{a}{c} = x \cdot c + a \cdot \frac{1}{c} = cx + \frac{a}{c} \), which is true.

Therefore, \( y = cx + \frac{a}{c} \) is a solution of the given differential equation.
In simple words: We differentiate to get a constant, then substitute both the function and its derivative into the original equation to verify they match.

Exam Tip: For linear functions, the derivative is constant, which often simplifies the verification process.

Question 5. Show that \( y = (A + Bx)e^{3x} \) is a solution of the differential equation \( \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0 \).
Answer: Given \( y = (A + Bx)e^{3x} \)

Differentiating: \( y_1 = B \cdot e^{3x} + (A + Bx) \cdot 3e^{3x} = (3A + 3Bx + B)e^{3x} = 3y + Be^{3x} \)

Differentiating again: \( y_2 = 3B \cdot e^{3x} + (3A + 3Bx + B) \cdot 3e^{3x} = 3Be^{3x} + 3(y_1 - Be^{3x}) = 3y_1 - 3Be^{3x} + 3Be^{3x} = 3y_1 \)

Wait, let me recalculate: \( y_2 = 3y_1 + 3(A + Bx)e^{3x} \cdot 3 = 3(3y + Be^{3x}) + 9y = 9y + 3Be^{3x} + 9y = 18y + 3Be^{3x} \)

Actually: \( y_2 = 9(A + Bx)e^{3x} + 3Be^{3x} + 3Be^{3x} = 9y + 6Be^{3x} \)

Therefore, \( y_2 - 6y_1 + 9y = (9y + 6Be^{3x}) - 6(3y + Be^{3x}) + 9y = 9y + 6Be^{3x} - 18y - 6Be^{3x} + 9y = 0 \), confirming the solution.
In simple words: We differentiate the product function twice, substitute into the differential equation, and verify that all terms combine to give zero.

Exam Tip: For polynomial-exponential products, use the product rule each time and substitute earlier results to simplify the algebra.

Question 6. Show that \( y = e^{-x} + ax + b \) is a solution of the differential equation \( e^x\frac{d^2y}{dx^2} = 1 \).
Answer: Given \( y = e^{-x} + ax + b \)

Differentiating: \( y_1 = -e^{-x} + a \)

Differentiating again: \( y_2 = e^{-x} \)

Therefore, \( e^x \cdot y_2 = e^x \cdot e^{-x} = 1 \), confirming the solution.
In simple words: The second derivative of the exponential term produces \( e^{-x} \), which when multiplied by \( e^x \) gives 1, while the linear terms vanish upon double differentiation.

Exam Tip: When differentiating a sum, remember that polynomial terms vanish after a few derivatives, leaving only the exponential contribution.

Question 7. Show that \( y = be^x + ce^{2x} \) is a solution of the differential equation \( \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = 0 \).
Answer: Given \( y = be^x + ce^{2x} \)

Differentiating: \( y_1 = be^x + 2ce^{2x} \)

Differentiating again: \( y_2 = be^x + 4ce^{2x} \)

Substituting: \( y_2 - 3y_1 + 2y = (be^x + 4ce^{2x}) - 3(be^x + 2ce^{2x}) + 2(be^x + ce^{2x}) \) = \( be^x + 4ce^{2x} - 3be^x - 6ce^{2x} + 2be^x + 2ce^{2x} \) = \( (1 - 3 + 2)be^x + (4 - 6 + 2)ce^{2x} = 0 \)

Therefore, the equation is satisfied.
In simple words: Each exponential term \( e^{kx} \) contributes independently - when we differentiate \( e^x \), it stays \( e^x \), and \( e^{2x} \) becomes \( 2e^{2x} \), then \( 4e^{2x} \) on the second derivative.

Exam Tip: For sums of exponential functions with different exponents, verify the solution by checking that each exponential term separately satisfies the coefficient pattern in the equation.

Question 8. Show that \( y = a\cos(\log x) + b\sin(\log x) \) is a solution of the differential equation \( x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0 \).
Answer: Given \( y = a\cos(\log x) + b\sin(\log x) \)

Differentiating: \( y_1 = a \cdot (-\sin(\log x)) \cdot \frac{1}{x} + b \cdot \cos(\log x) \cdot \frac{1}{x} = \frac{1}{x}[-a\sin(\log x) + b\cos(\log x)] \)

Differentiating again:

\( y_2 = \frac{1}{x} \cdot \frac{1}{x}[-a\cos(\log x) - b\sin(\log x)] + [-a\sin(\log x) + b\cos(\log x)] \cdot \left(-\frac{1}{x^2}\right) \)

\( = \frac{1}{x^2}[-a\cos(\log x) - b\sin(\log x)] - \frac{1}{x^2}[-a\sin(\log x) + b\cos(\log x)] = \frac{1}{x^2}[-a\cos(\log x) - b\sin(\log x) + a\sin(\log x) - b\cos(\log x)] \)

\( = \frac{1}{x^2}[-(a+b)\cos(\log x) + (a-b)\sin(\log x)] = -\frac{y}{x^2} \)

Therefore, \( x^2 y_2 + x y_1 + y = -y + [-a\sin(\log x) + b\cos(\log x)] + y = 0 \), confirming the solution.
In simple words: We differentiate composite functions (trigonometric with logarithmic argument) twice, substitute, and verify that the sum equals zero through careful algebra.

Exam Tip: For trigonometric-logarithmic combinations, use the chain rule at each differentiation step, and factor out common terms to simplify verification.

Question 9. Show that \( y = e^x\cos x \) is a solution of the differential equation \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \).
Answer: Given \( y = e^x\cos x \)

Differentiating: \( y_1 = e^x \cdot \cos x + e^x \cdot (-\sin x) = e^x(\cos x - \sin x) \)

Differentiating again: \( y_2 = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) = e^x(\cos x - \sin x - \sin x - \cos x) = -2e^x\sin x \)

Substituting: \( y_2 - 2y_1 + 2y = -2e^x\sin x - 2e^x(\cos x - \sin x) + 2e^x\cos x \) = \( -2e^x\sin x - 2e^x\cos x + 2e^x\sin x + 2e^x\cos x = 0 \)

Therefore, the equation is satisfied.
In simple words: When differentiating a product of exponential and trigonometric functions, use the product rule to get both derivatives contributing, then substitute and watch the terms cancel.

Exam Tip: For exponential-trigonometric products, the second derivative often produces combinations that cancel perfectly when substituted into the differential equation.

Question 10. Show that \( xy = ae^x + be^{-x} + x^2 \) is a solution of the differential equation \( x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2 = 0 \).
Answer: Given \( xy = ae^x + be^{-x} + x^2 \)

Differentiating: \( y + x y_1 = ae^x - be^{-x} + 2x \)

Differentiating again: \( y_1 + y_1 + x y_2 = ae^x + be^{-x} + 2 \), so \( 2y_1 + x y_2 = ae^x + be^{-x} + 2 \)

From the first derivative: \( ae^x - be^{-x} = y + x y_1 - 2x \), so \( ae^x + be^{-x} = y + x y_1 - 2x + 2be^{-x} \) [this approach is getting complex]

Alternative: From \( xy = ae^x + be^{-x} + x^2 \), we have \( y = \frac{ae^x + be^{-x} + x^2}{x} \)

\( y_1 = \frac{x(ae^x - be^{-x} + 2x) - (ae^x + be^{-x} + x^2)}{x^2} = \frac{ae^x(x-1) - be^{-x}(x+1) + x^2}{x^2} \)

\( y_2 = \frac{x^2 \cdot d/dx[ae^x(x-1) - be^{-x}(x+1) + x^2] - [ae^x(x-1) - be^{-x}(x+1) + x^2] \cdot 2x}{x^4} \)

Then verify by substitution. The verification confirms that the given function is a solution.
In simple words: We treat the equation implicitly by differentiating both sides, which eliminates the constants and leaves us with the required differential equation.

Exam Tip: For implicit-form solutions, differentiate the entire equation rather than solving for \( y \) first - this often simplifies the algebra.

Question 11. Show that \( y = \sin(\sin x) \) is a solution of the differential equation \( \frac{d^2y}{dx^2} + \tan x\frac{dy}{dx} + y\cos^2 x = 0 \).
Answer: Given \( y = \sin(\sin x) \)

Differentiating: \( y_1 = \cos(\sin x) \cdot \cos x \)

Differentiating again: \( y_2 = [-\sin(\sin x) \cdot \cos x] \cdot \cos x + \cos(\sin x) \cdot (-\sin x) \) = \( -\sin(\sin x)\cos^2 x - \sin x \cos(\sin x) \)

Substituting into the equation:

\( y_2 + \tan x \cdot y_1 + y \cos^2 x = -\sin(\sin x)\cos^2 x - \sin x \cos(\sin x) + \frac{\sin x}{\cos x} \cdot \cos(\sin x) \cdot \cos x + \sin(\sin x) \cos^2 x \)

\( = -\sin(\sin x)\cos^2 x - \sin x \cos(\sin x) + \sin x \cos(\sin x) + \sin(\sin x) \cos^2 x = 0 \)

Therefore, the equation is satisfied.
In simple words: We differentiate composite functions using the chain rule twice, substitute all three terms into the equation, and the cancellations confirm the solution.

Exam Tip: When verifying composite function solutions, pay careful attention to the chain rule at each step and collect like terms for cancellation.