Laplace transform #12 - Q&A

Question 1 Find the solution of the following differential equations: Homogeneous Differential Equations 3xy²dx + (3x²y + y³) dy = 0; y(1) = 2 (x²-9y²)dx + 2xydy = 0; y(2) = -1 Answer The given homogenous differential equation 3xy²dx + (3x²y + y³) dy = 0 (1) y(1) = 2 (2) First we check the Exactness of the given Differential equation, Mdx+Ndy = 0 M = 3xy², N = 3x²y + y³ ∂M/∂y = 6xy, ∂N/∂x = 6xy ∂M/∂y = ∂N/∂x So, the given differential equation is exact. Hence the solution is given by. ∫Mdx + ∫(terms in N not containg x)dy = c ∫3xy²dx + ∫y³dy = c 3x²y²/2 + y⁴/4 = c 3x²y²/2 + y⁴/4 = c (3) using (2) Put x = 1, y = 2 3(1)²(2)²/2 + (2)⁴/4 = c 6+4= c => c = 10 (3) becomes 3x²y²/2 + y⁴/4 = 10 3x²y² + y⁴ = 40 This is the required solution. Answer 2 (x²-9y²)dx + 2xydy = 0 y(2) = -1 (x² - 9y²)dx = -2xydy dy/dx = -(x²-9y²)/2xy y = vx dy/dx = v + xdv/dx v + xdv/dx = -(x² - 9v²x²)/(2xv) v + xdv/dx = -(1 - 9v²)/(2v) x dv/dx = -(1 - 9v²)/(2v) - v x dv/dx = (-1 + 9v² - 2v²)/2v x dv/dx = (7v² - 1)/2v ∫2v/(7v² - 1) dv = ∫1/x dx let 7v² - 1 = t => 14v dv = dt ∫(1/7t) dt = ∫1/x dx 1/7 ln(t) = ln(x) + lnc 1/7 ln(7v² - 1) = ln(xc) ln(7v² - 1) = 7ln(xc) 7v² - 1 = (xc)^7 (7(y²/x²) - 1) = (xc)^7 Question 2 Solve the following manually using separable differential equations method. mydx + nxdy = 0 Answer 1 mydx + nxdy = 0 mydx = -nxdy dy/dx = -m/nx/y Take both side integration: ∫dx/x = ∫-n/mdy/y ln(x) = -n/m ln(y) + c (.. c = const) Answer: ln (x) + n/m ln(y) = c Answer 2 Simplifying mydx + nxdy = 0 Solving dmxy + -1dnxy = 0 Solving for variable 'd'. Move all terms containing d to the left, all other terms to the right. Factor out the Greatest Common Factor (GCF), 'dxy'. dxy(m + -1n) = 0 Subproblem 1 Set the factor 'dxy' equal to zero and attempt to solve: Simplifying dxy = 0 Solving dxy = 0 Move all terms containing d to the left, all other terms to the right. Simplifying dxy = 0 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined. Subproblem 2 Set the factor'(m + -1n)' equal to zero and attempt to solve: Simplifying m + -1n = 0 Solving m + -1n = 0 Move all terms containing d to the left, all other terms to the right. Add '-1m' to each side of the equation. m + -1n + -1m = 0 + -1m Combine like terms: m + -1m = 0 0 + -1n = 0 + -1m -1n = 0 + -1m Remove the zero: -1n = -1m Add 'n' to each side of the equation. -1n + n = -1m + n Combine like terms: -1n + n = 0 0 = -1m + n Simplifying 0 = -1m + n The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined. The solution to this equation could not be determined. Answer 3 mydx = nxdy ⇒ my dy = nx dx Substitute dy/dx with y'. ⇒ my = nx y' y/y' = nx/m ⇒ 1/y dy = m/n dx/x ⇒ ∫1/y dy = m/n∫1/x dx + C1 ⇒ ln y = m/n ln(x) + C1 Answer: y = e^(m/n ln(x)) + C1 Question 3 Find the orthogonal trajectory of the family of curves y = cx². Answer 1 y = cx² dy/dx = 2cx c = (1/2x)dy/dx y = (1/2x)dy/dx * x² y = x/2 (dy/dx) Replacing dy/dx with -(dx/dy) y = x/2 (-dx/dy) y(dy/dx) = -x/2 ∫y dy = -∫x/2 dx y²/2 = -x²/4 + A²/4 2y² = -x² + A² Answer x² + 2y² = A² where A is constant. Answer 2 y = Cx² ⇒ y/x² = C, Differentiating (x²y' - 2xy) / x⁴ = 0 ⇒ y' = 2y/x Replacing dy/dx by -dx/dy, we have y' = -x/2y ⇒ 2ydy + xdx = 0 ⇒ y²/2 + x²/4 = constant which represent a family of ellipses. Answer 3 y = cx² (1) differentiating (1) w,r,t x, we get y' = 2cx (2) From (1) c = y/x², using in (2), we get y' = 2(y/x²)x = 2y/x (3) Since, slopes of arthogonal trajectories are negative reciprocals of each other, then orthogonal trajectory is given by solving: y' = -x/2y → Slope of orthogonal (4) trajectories Solving (4), dy/dx = -x/2y ⇒ 2y dy = -xdx ⇒ ∫2ydy = -∫xdx ⇒ y² = -x²/2 + K1 or 2y² + x² = K; K = 2K1 So (5) represent arthogonal trajectories of (1) Question 4 A metal is heated up to a temperature of 500°C. It is then exposed to a temperature of 38°C. After 2 minutes, the temperature of the metal becomes 190°C. When will the temperature be 100°C? Answer 1 Initial temp = 500°C Ta = Ambient temp = 38°C After 2 min, temp = 190°C Calculate: Time when temp reaches to 100°C dT/dt = -K(T-Ta) dT/(T-Ta) = -K dt ∫(190 to 500) dT/(T-Ta) = -K ∫(0 to 2)dt t = 0, T= 500°C t = 2, T = 190°C [ln (T-Ta)] from 500 to 190 = -K(2 – 0) ln (190-38) - ln(500-38) = -2K -ln (231/76) = -2K K = 1/2 ln(231/76) dT/(T-Ta) = -Kdt t = 0, T= 500°C T=t, T = 100°C ∫(100 to 500) dT/(T-Ta) = -k ∫(0 to t) dt [ln (T-Ta)] from 500 to 100 = -k(t-0) ln(100-38) - ln(500-38) = -kt ln(62/462) = -kt t = ln(462/62) / (1/2 ln(231/76)) t = (ln(462/62)) / (1/2 ln(231/76)) t = 3.61 minute Answer: t = 3.61 minute Answer 2 The mean temperature of the metal as it cools from 500°C to 190°C is, θ = (500+190)/2 θ = 690/2 = 345°C The rate of decrease of temperature is, (500 - 190)/2 = 155 deg C/min According to the Newtons Answer 3 Ta = Ambient temp = 38°C dT/dt = -K(T - Ta) = 190°C = 100°C dT/(T-Ta) = -K dt ∫(190 to 500) dT/(T-Ta) = -K∫(0 to 2)dt t = 0, T = 500°C t = 2, T = 190°C [ln(T-Ta)] from 500 to 190 = -K(2-0) ln(190-38) - ln(500-38) = -2K -ln (231/76) = -2K K = 1/2 ln(231/76) ∫(100 to 500) dT/(T-Ta) = -k ∫(0 to t)dt t = 0, T = 500°C T = t, T = 100°C [ln(T-Ta)] from 500 to 100 = -k(t-0) ln(100-38) - ln(500-38) = -kt ln(62/462) = -1/2 ln(231/76)t Answer: t = 3.61 minute Question 5 Auxiliary Equation: Solve the following: (3D³ – 22D² +7D)y = 0 Answer 1 A.E. (Auxiliary equation) 3m³ - 22m² + 7m = 0 m(3m² - 22m + 7) = 0 m = 0,3m² - 21m -m+7=0 3m(m-7) - 1(m – 7) = 0 (3m-1)(m-7) = 0 m = 7, m = 1/3 Then m = 0, m = 7, m = 1/3 y = c₁e⁰t + c₂e^(7t) + c₃e^( t/3 ) y = C1 + c₂e^(7z) + c₃e^(z/3) Answer 2 To Solve: The above auxiliary equation. Explanation- Rewrite the given expression, (3D³ – 22D² + 7D)y = 0 Solving as follows by factorising the above equation , we get, D(3D² – 22D + 7) = 0 D((3D² – 21D – D + 7)) = 0 D(3D(D-7) - 1(D – 7)) = 0 D(3D-1)(D-7) = 0 Further solving, we get, D = 0, 1/3 and 7 So, the complementry equation can be written as follows, C. E. = C₁e⁰ + C₂e^( t/3 ) + C₃e^(7t) = C₁ + C₂e^( t/3 ) + C₃e^(7t) Answer: Hence, the solution of the auxiliary equation (3D³ – 22D² + 7D)y = 0 is C₁ + C₂e^( t/3 ) + C₃e^(7t) Answer 3 (3D³ – 22D² + 7D)y = 0 D(3D² - 22D + 7) = 0 D(3D² – 21D – D + 7) = 0 D(3D(D-7) - 1(D – 7)) = 0 D(3D-1)(D-7) = 0 D = 0, 1/3 and 7K C. E. = C₁e⁰ + C₂e^( t/3 ) + C₃e^(7t) = C1 + C2e^( t/3 ) + C3e^(7t) Answer: (3D³ – 22D² + 7D)y = 0 is C1+C2e^( t/3 ) + C3e^(7t)