Differential equations #1 - Questions and Answers

Question 1 Find the general solution of the given higher-order differential equation. y⁽⁴⁾ + y''' + y'' = 0 y(x) = Answer The characteristic equation is r⁴ + r³ + r² = 0 r⁴ + r³ + r² = 0 r²(r² + r + 1) = 0 Observe that r = 0 is a root with multiplicity 2. r = (-1 ± √((-1)² - 4(1)(1)) ) / (2(1)) r = (-1 ± √(1 - 4) ) / 2 r = (-1 ± √(-3) ) / 2 r = (-1 ± i√3 ) / 2 The answer y(x) = c₁ + c₂x + c₃e^(-x/2)cos(√3/2 x) + c₄e^(-x/2)sin(√3/2 x) Question 2 Solve the given differential equation by separation of variables: dy/dx = x√(1 - y²) Answer Given that differential equation dy/dx = x√(1 - y²) Applying integration on both sides. ∫ dy/√(1 - y²) = ∫ x dx sin⁻¹(y) = x²/2 + C [∫ du/√(a²-u²) = sin⁻¹(u/a)] y(x) = sin(x²/2 + C) Question 3 If xy + beʸ = be, find the value of y'' at the point where x=0 Answer Solution: - We have given xy + beʸ = be for x=0. 0(y) + beʸ = be => beʸ = be => eʸ = e => y=1 Hence, we have y=1 for x=0 now, xy + beʸ = be differentiate both sides, we get y + xy' + beʸy' = 0 -- ① for x=0, y=1, we have 1 + 0(y') + be¹y' = 0 => 1 + bey' = 0 => y' = -1/(be) differentiate both sides, we get y'' + y' + xy'' + beʸy'² + bey'' = 0 Plugging valve of x, y, y', we get -1/(be) + 0(y'') + be*(-1/(be))² + bey'' = 0 => -1/(be) - 1/(be) + bey'' + 1/(be) = 0 => -2/(be) + (1/be) + bey'' = 0 => 1/(be) = -bey'' => y'' = -1/(b²e²) Question 4 Solve the equation explicitly for y and differentiate to get y' in terms of x. 1/x + 1/y = 1 Answer 1 1/x + 1/y = 1 1/y = 1 - 1/x 1/y = (x-1)/x Take reciprocal of both sides y = x/(x-1) Differentiate both sides with respect to x dy/dx = d/dx(x/(x-1)) dy/dx = ((x - 1) * 1 - x1)/((x-1)²) [QUOTIENT RULE (u/v)’ = (vu’ - uv’) / v² ] dy/dx = (x - 1 - x)/(x-1)² dy/dx = -1/(x-1)² Result: dy/dx = -1/(x-1)² Answer 2 To solve the equation 1/x + 1/y = 1 explicitly for y and differentiate to get y' in terms of x, we can follow these steps: Step 1: Rearrange the equation to isolate y: 1/y = 1 - 1/x Step 2: Take the reciprocal of both sides: y = 1 / (1 - 1/x) Step 3: Simplify the expression on the right-hand side by finding a common denominator: y = 1 / ((x-1)/x) = x/(x-1) Step 4: Invert the fraction and multiply: y = x/(x-1) Now, to differentiate y with respect to x, we can use the quotient rule. Let's denote the derivative of y with y': y = x / (x-1) Using the quotient rule, we have: y' = ((x-1)d/dx(x) - xd/dx(x-1)) / (x-1)² Simplifying this expression gives us: y' = ((x-1)(1) - x(1)) / (x-1)² y' = (x - 1 - x) / (x-1)² y' = -1 / (x-1)² Further simplification yields: y' = -1/(x-1)² Hence, the solution to the equation and the derivative y' are: y = x / (x-1) y' = -1/(x-1)² Answer 3 Step 1: Begin by subtracting 1/x from both sides of the equation: 1/y = 1 - 1/x Step 2: Take the reciprocal of both sides to isolate y: y = 1 / (1 - 1/x) Step 3: Simplify the expression on the right side by finding a common denominator: y = 1 / ((x-1)/x) y = x / (x - 1) Therefore, the equation explicitly solved for y is: y = x / (x - 1) Step 4: To find y', we need to differentiate y with respect to x. Using the quotient rule, we have: y' = d/dx (x/(x-1)) = ((x-1)d/dx(x) - xd/dx(x-1)) / (x-1)² Simplifying further: y' = ((x-1)(1) - x(1)) / (x-1)² y' = (x - 1 - x) / (x-1)² = -1 / (x-1)² Hence, y' in terms of x is: y' = -1 / (x-1)² Answer 4 Step 1: Solving for y We'll start by rearranging the equation to isolate y on one side: 1/y = 1 - 1/x Next, we'll take the reciprocal of both sides to solve for y: y = 1 / (1 - 1/x) Step 2: Differentiating to find y' in terms of x To differentiate y with respect to x, we'll use the chain rule. Let's denote 1/(1- 1/x) as u, and we'll find du/dx u = 1/(1 - 1/x) To differentiate u, we'll apply the chain rule: du/dx = -1/(1-1/x)² * d/dx(1 - 1/x) Using the quotient rule, the derivative of 1/x can be found as: d/dx(1/x) = (x0 - 11)/x² = -1/x² The derivative of (1 - 1/x) can be calculated as: d/dx(1 - 1/x) = 0 - (-1/x²) = 1/x² Plugging this value back into the equation, we have: du/dx = -1/(1-1/x)² * (1/x²) Simplifying further, we get: du/dx = 1 / (x²(1-1/x)²) Finally, substituting back u as 1/(1-1/x), we obtain: y' = 1 / (x²(1 - 1/x)²) Therefore, the derivative y' in terms of x is given by: y' = -1 / (x-1)² To summarize, the solution to the equation 1/x + 1/y = 1 is given by y = x/(x-1) and the derivative of y with respect to x is y' = -1/(x-1)² Question 5 Find transient terms in this general solution to a differential equation, if there are any y = (x + C) (x/(x - 2)) Answer 1 y = (x+C) (x+2)/(x-2) = (x+C) (x+2) / (x-2) = (x² + (C+2)x + 2C) / (x - 2) = x + (2+C) + 2C/(x-2) Then take the limits of each of these three terms as x → ∞ and see which (if any) approach 0. All three are rational expressions, so it is clear that only 2C/(x-2) is transient term. Answer 2 To find the transient terms in the general solution of the given differential equation, we first need to rewrite the equation in a more standard form. The given general solution is: y = (x + C) (x / (x-2)) To simplify the expression, we can expand it: y = (x(x+2) + C(x+2)) / (x-2) = (x² + Cx + 2x + 2C) / (x-2) = (x² + (C+2)x + 2C) / (x-2) Now, let's write the expression in a more suitable form for identifying the transient terms: y = (x² + (C+2)x + 2C) / (x - 2) The transient terms are the terms in the numerator that have a power of x greater than or equal to the power in the denominator. In this case, the highest power in the denominator is 1 (since it is linear in x). Therefore, we are interested in the terms with x raised to the power of 2 or higher. The numerator can be expressed as: x² + (C+2)x + 2C The terms x² and Cx have powers of x greater than 1. Thus, these are the transient terms. Therefore, the transient terms in the general solution are x² and Cx. Answer 3 y = (x + C) (x / (x - 2)) To analyze the transient terms, we'll rewrite the expression inside the parentheses as: x/(x-2) = (x-2+2) / (x-2) = 1 + 2/(x-2) Now we can substitute this back into the original equation: y = (x + C) (1 + 2/(x-2)) Expanding this expression, we get: y = x + C + (2x + 2C)/(x-2) Now we can identify the transient terms. The transient terms are those that tend to zero as x goes to infinity. In this case, the term (2x + 2C)/(x - 2) tends to zero as x approaches infinity, since the numerator grows linearly while the denominator grows faster. Therefore, the transient terms in the general solution are: (2x+2C)/(x-2)