Differential Equations Ex 9.6: Linear First Order Equations

Understanding Linear Differential Equations of the First Order

A differential equation is called linear if the dependent variable and its derivatives appear only in the first degree, and they are not multiplied together. In Exercise 9.6, we focus on first-order linear differential equations, which can be expressed in one of two standard forms:

1. Form 1: dy/dx + Py = Q, where P and Q are functions of 'x' or constants.
2. Form 2: dx/dy + Px = Q, where P and Q are functions of 'y' or constants.

The key to solving these equations lies in a special function called the 'Integrating Factor' (IF). The integrating factor transforms the left-hand side of the equation into the derivative of a product, making the equation directly integrable. This method elegantly simplifies a seemingly complex differential equation into a form that can be solved using basic integration techniques. Identifying P and Q correctly is the first critical step, followed by calculating the integrating factor, and then performing the integration to find the general solution. This systematic approach ensures accurate solutions for a wide range of problems in this category.

Key Definitions

Linear Differential Equation

A differential equation is linear if the dependent variable (y) and its derivatives (dy/dx, d²y/dx², etc.) appear only in the first degree, and there are no products of the dependent variable with its derivatives.

First Order Linear Differential Equation

A linear differential equation involving only the first derivative, typically written in the form dy/dx + Py = Q or dx/dy + Px = Q.

Integrating Factor (IF)

A function that, when multiplied by a linear differential equation, makes the left-hand side an exact derivative. For dy/dx + Py = Q, IF = e^(∫Pdx). For dx/dy + Px = Q, IF = e^(∫Pdy).

General Solution

The solution to a differential equation that includes an arbitrary constant (C), representing a family of curves that satisfy the equation.

Step-by-Step Method to Solve Linear First Order Differential Equations

1. Step 1: Identify the Standard Form — Examine the given differential equation and rewrite it into one of the standard linear forms: dy/dx + Py = Q (if 'P' and 'Q' are functions of x) or dx/dy + Px = Q (if 'P' and 'Q' are functions of y). Clearly identify P and Q.
2. Step 2: Calculate the Integrating Factor (IF) — Once P is identified: If the form is dy/dx + Py = Q, then IF = e^(∫Pdx). If the form is dx/dy + Px = Q, then IF = e^(∫Pdy). Remember to perform the integration for ∫Pdx or ∫Pdy and then raise 'e' to that power. Do not include the constant of integration 'C' when calculating IF.
3. Step 3: Apply the Solution Formula — Multiply the entire standard form equation by the Integrating Factor. Alternatively, use the direct solution formula: For dy/dx + Py = Q: y (IF) = ∫(Q IF)dx + C For dx/dy + Px = Q: x (IF) = ∫(Q IF)dy + C This formula is derived by integrating both sides after multiplying by the IF.
4. Step 4: Integrate the Right Hand Side — Carefully integrate the product (Q * IF) with respect to 'x' (for the first form) or 'y' (for the second form). This step often requires knowledge of various integration techniques like integration by parts, substitution, etc. Don't forget to add the constant of integration 'C' at this stage.
5. Step 5: Simplify and Find the General Solution — Isolate the dependent variable (y or x) to express the general solution. The general solution will contain the arbitrary constant 'C'. If an initial condition is given, substitute it into the general solution to find the particular solution (value of C).

Worked Examples

• Example 1: Solve dy/dx + y/x = x² Step 1: Identify the standard form. This is in the form dy/dx + Py = Q, where P = 1/x and Q = x². Step 2: Calculate the Integrating Factor (IF). IF = e^(∫Pdx) = e^(∫(1/x)dx) = e^(ln|x|) = x (assuming x > 0). Step 3: Apply the solution formula. y (IF) = ∫(Q IF)dx + C y x = ∫(x² x)dx + C xy = ∫x³dx + C Step 4: Integrate the Right Hand Side. xy = x⁴/4 + C Step 5: Simplify and find the general solution. y = (x³/4) + C/x Example 2: Solve (1 + y²)dx + (x - e^(arctan y))dy = 0 Step 1: Identify the standard form. First, rearrange the equation to resemble dx/dy + Px = Q. (1 + y²)dx = -(x - e^(arctan y))dy dx/dy = -(x - e^(arctan y))/(1 + y²) dx/dy = -x/(1 + y²) + e^(arctan y)/(1 + y²) dx/dy + x/(1 + y²) = e^(arctan y)/(1 + y²) This is in the form dx/dy + Px = Q, where P = 1/(1 + y²) and Q = e^(arctan y)/(1 + y²). Step 2: Calculate the Integrating Factor (IF). IF = e^(∫Pdy) = e^(∫(1/(1+y²))dy) = e^(arctan y). Step 3: Apply the solution formula. x (IF) = ∫(Q IF)dy + C x e^(arctan y) = ∫( (e^(arctan y)/(1 + y²)) e^(arctan y) )dy + C x e^(arctan y) = ∫( (e^(2arctan y))/(1 + y²) )dy + C Step 4: Integrate the Right Hand Side. Let t = arctan y. Then dt = 1/(1 + y²) dy. So, ∫( (e^(2arctan y))/(1 + y²) )dy becomes ∫e^(2t)dt. ∫e^(2t)dt = (1/2)e^(2t) + C Substitute t back: (1/2)e^(2arctan y) + C Thus, x e^(arctan y) = (1/2)e^(2arctan y) + C Step 5: Simplify and find the general solution. x = (1/2)e^(arctan y) + C e^(-arctan y) Example 3: Find the particular solution of x dy/dx + 2y = x² log x, given that y = 1 when x = 1. Step 1: Identify the standard form. Divide by x: dy/dx + (2/x)y = x log x. This is in the form dy/dx + Py = Q, where P = 2/x and Q = x log x. Step 2: Calculate the Integrating Factor (IF). IF = e^(∫(2/x)dx) = e^(2ln|x|) = e^(ln|x|²) = x². Step 3: Apply the solution formula. y (IF) = ∫(Q IF)dx + C y x² = ∫(x log x * x²)dx + C yx² = ∫x³ log x dx + C Step 4: Integrate the Right Hand Side (using Integration by Parts: ∫u dv = uv - ∫v du). Let u = log x, dv = x³ dx. Then du = (1/x)dx, v = x⁴/4. ∫x³ log x dx = (log x)(x⁴/4) - ∫(x⁴/4)(1/x)dx = (x⁴/4)log x - ∫(x³/4)dx = (x⁴/4)log x - x⁴/16 So, yx² = (x⁴/4)log x - x⁴/16 + C Step 5: Find the particular solution. Given y = 1 when x = 1: (1)(1)² = (1⁴/4)log(1) - 1⁴/16 + C 1 = (1/4)(0) - 1/16 + C 1 = -1/16 + C C = 1 + 1/16 = 17/16 Substitute C back into the general solution: yx² = (x⁴/4)log x - x⁴/16 + 17/16 y = (x²/4)log x - x²/16 + 17/(16x²) This is the particular solution.

Exam Tip: Avoiding Common Pitfalls

When solving linear differential equations in your CBSE exams, pay close attention to these details to avoid losing marks:

1. Correct Standard Form: Always ensure the equation is precisely in the dy/dx + Py = Q or dx/dy + Px = Q form. Missing coefficients or incorrect signs can lead to errors in P and Q.
2. Identifying P and Q: P and Q must be solely functions of 'x' for dy/dx form, or solely functions of 'y' for dx/dy form. If a term involves both x and y, you might need to reconsider the form or if it's indeed linear.
3. Integrating Factor Calculation: The integral ∫Pdx or ∫Pdy must be calculated accurately. Remember ∫(1/x)dx = ln|x| and e^(ln|f(x)|) = f(x). Do not add 'C' when calculating the IF.
4. Integration of RHS: The integral ∫(Q IF)dx or ∫(Q IF)dy can be complex. Be thorough with your integration techniques (substitution, by parts, standard formulas). This is where you must add the constant of integration 'C'.
5. Final Simplification: Isolate 'y' or 'x' to present the solution clearly. If a particular solution is required, use the given initial conditions to find the value of 'C' correctly.

Practice Questions with Solutions

• Q: Solve the differential equation: (x + 2y³)dy/dx = y A: Step 1: Rewrite the equation in standard form. The given equation is (x + 2y³)dy/dx = y. It is easier to write it in the form dx/dy + Px = Q. dx/dy = (x + 2y³)/y dx/dy = x/y + 2y² dx/dy - (1/y)x = 2y² Here, P = -1/y and Q = 2y². Step 2: Calculate the Integrating Factor (IF). IF = e^(∫Pdy) = e^(∫(-1/y)dy) = e^(-ln|y|) = e^(ln|y⁻¹|) = y⁻¹ = 1/y. Step 3: Apply the solution formula. x (IF) = ∫(Q IF)dy + C x (1/y) = ∫(2y² (1/y))dy + C x/y = ∫2y dy + C Step 4: Integrate the Right Hand Side. x/y = 2(y²/2) + C x/y = y² + C Step 5: Simplify to find the general solution. x = y³ + Cy Final answer: x = y³ + Cy
• Q: Find the general solution of the differential equation: cos²x dy/dx + y = tan x (0 < x < π/2) A: Step 1: Rewrite the equation in standard form. Divide by cos²x: dy/dx + (1/cos²x)y = tan x / cos²x dy/dx + (sec²x)y = tan x sec²x Here, P = sec²x and Q = tan x sec²x. Step 2: Calculate the Integrating Factor (IF). IF = e^(∫Pdx) = e^(∫sec²x dx) = e^(tan x). Step 3: Apply the solution formula. y (IF) = ∫(Q IF)dx + C y e^(tan x) = ∫(tan x sec²x e^(tan x))dx + C Step 4: Integrate the Right Hand Side. Let t = tan x, so dt = sec²x dx. The integral becomes ∫t e^t dt. Using integration by parts (∫u dv = uv - ∫v du): Let u = t, dv = e^t dt. Then du = dt, v = e^t. ∫t e^t dt = t e^t - ∫e^t dt = t e^t - e^t = e^t (t - 1). Substitute t = tan x back: e^(tan x) (tan x - 1). So, y e^(tan x) = e^(tan x) (tan x - 1) + C Step 5: Simplify to find the general solution. Divide by e^(tan x): y = (tan x - 1) + C e^(-tan x) Final answer: y = (tan x - 1) + Ce^(-tan x)
• Q: Solve the differential equation: x log x dy/dx + y = 2/x log x A: Step 1: Rewrite the equation in standard form. Divide by x log x: dy/dx + y/(x log x) = (2/x log x) / (x log x) dy/dx + (1/(x log x))y = 2/(x² log x) Here, P = 1/(x log x) and Q = 2/(x² log x). Step 2: Calculate the Integrating Factor (IF). IF = e^(∫(1/(x log x))dx) Let t = log x, so dt = (1/x)dx. ∫(1/(x log x))dx = ∫(1/t)dt = ln|t| = ln|log x|. So, IF = e^(ln|log x|) = log x (assuming x > 0 and log x > 0). Step 3: Apply the solution formula. y (IF) = ∫(Q IF)dx + C y log x = ∫( (2/(x² log x)) log x )dx + C y log x = ∫(2/x²)dx + C Step 4: Integrate the Right Hand Side. y log x = ∫2x⁻²dx + C y log x = 2 * (x⁻¹/(-1)) + C y log x = -2/x + C Step 5: Simplify to find the general solution. y = (-2/x log x) + (C/log x) Final answer: y = C/log x - 2/(x log x)
• Q: Find the particular solution of the differential equation (1 + x²)dy + 2xy dx = cot x dx, given that y = 0 when x = π/2. A: Step 1: Rewrite the equation in standard form. Divide by dx: (1 + x²)dy/dx + 2xy = cot x Divide by (1 + x²): dy/dx + (2x/(1 + x²))y = cot x / (1 + x²) Here, P = 2x/(1 + x²) and Q = cot x / (1 + x²). Step 2: Calculate the Integrating Factor (IF). IF = e^(∫Pdx) = e^(∫(2x/(1+x²))dx) Let t = 1 + x², so dt = 2x dx. ∫(2x/(1+x²))dx = ∫(1/t)dt = ln|t| = ln|(1 + x²)|. So, IF = e^(ln(1+x²)) = 1 + x² (since 1+x² is always positive). Step 3: Apply the solution formula. y (IF) = ∫(Q IF)dx + C y (1 + x²) = ∫( (cot x / (1 + x²)) (1 + x²) )dx + C y(1 + x²) = ∫cot x dx + C Step 4: Integrate the Right Hand Side. y(1 + x²) = ln|sin x| + C Step 5: Find the particular solution using y = 0 when x = π/2. 0 * (1 + (π/2)²) = ln|sin(π/2)| + C 0 = ln|1| + C 0 = 0 + C C = 0 Substitute C = 0 back into the general solution: y(1 + x²) = ln|sin x| y = ln|sin x| / (1 + x²) Final answer: y = ln|sin x| / (1 + x²)

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