Differential equations #2 - Questions and Answers

Question 1 Solve the given differential equation by separation of variables. csc ydx + sec²xdy = 0 Answer csc y dx + sec²x dy = 0 Rearrange terms sec²x dy = -csc y dx Separate the variables dy / csc y = -dx / sec²x sin y dy = -cos²x dx Integrate both sides ∫ sin y dy = - ∫ cos²x dx Use the identity cos²x = (1 + cos 2x)/2, SO ∫ sin y dy = - ∫ (1 + cos 2x)/2 dx Distribute ∫ sin y dy = - (1/2) ∫ dx - (1/2)∫ cos 2x dx Rewrite ∫ sin y dy = - (1/2) ∫ dx - (1/2)∫ cos 2x dx Apply ∫sin u du = -cosu + C and ∫cos u du = -sin u + C -cos y = - (1/2)x - (1/4)sin2x + C Multiply both sides by -4 4cos y = 2x + sin 2x + C Result: 4cos y = 2x + sin 2x + C Question 2 if tax revenue is $230 billion and the governments Answer The correct answer is e) deficit is 5 b i l l i o n a n d g o v e r n m e n t d e b t w i l l i n c r e a s e b y 5 5billionandgovernmentdebtwillincreaseby5 billion Question 3 Solve the differential equation by variation of parameters y'' + 3y' + 2y = 1 / (1 + eˣ) Answer Given: y'' + 3y' + 2y = 1/(1 + eˣ) --- (1) (1) is of the form: y''+P(x)y' + Q(x)y = R(x) Where, P(x) = 3; Q(x)=2, R(x)= 1/(1 + eˣ) To find: Homogeneous solution. Solution: Consider the homogeneous part of equation (1) i.e., y'' + 3y' + 2y = 0 The auxiliary equation is m² + 3m + 2 = 0 The factors are (m+2)(m+1)=0 The roots are m₁ = -2, m₂ = -1 ∴ y_c(x) = c₁e^(-2x) + c₂e^(-x) Obtain the particular solution. y_p(x) = A(x)y₁(x) + B(x)y₂(x) where, A(x) = -∫(y₂(x) R(x) / W(y₁(x), y₂(x))) dx B(x) = ∫(y₁(x) R(x) / W(y₁(x), y₂(x))) dx Here, y₁(x) = e^(-2x) , y₂(x) = e^(-x) The Wronskian (W(y₁, y₂)) is obtained as follows W(y₁, y₂) = | y₁(x) y₂(x) | = | e^(-2x) e^(-x) | = e^(-2x) * -e^(-x) - (-2e^(-2x))*e^(-x) = -e^(-3x) + 2e^(-3x) | y₁'(x) y₂'(x) | | -2e^(-2x) -e^(-x)| = e^(-3x) A(x) = -∫(y₂(x) R(x) / W(y₁(x), y₂(x))) dx = -∫(e^(-x) / (1 + eˣ) * 1/e^(-3x)) dx = -∫(e^(2x) / (1 + eˣ) )dx A(x) = -∫(e^(2x) / (1 + eˣ) )dx = - ∫(eˣ / (1 + eˣ) * (eˣ) dx = -∫(eˣ/(1 + eˣ))dx= -e^(-x) + ln(1 + e^x) -1 B(x) = ∫(y₁(x) R(x) / W(y₁(x), y₂(x))) dx = ∫(e^(-2x) / (1 + eˣ) * 1/e^(-3x) ) dx= ∫(e^(x) / (1 + eˣ) ) dx = ∫(eˣ / (1 + eˣ))dx = ln(1 + eˣ) y_p(x) = [ -e^(-x) + ln(1 + eˣ) -1] e^(-2x) + [ln(1 + eˣ)] e^(-x) y_p(x) = -e^(-3x) + e^(-2x)ln(1 + eˣ) - e^(-2x) + e^(-x)ln(1 + eˣ) y_p(x) = (-e^(-3x) - e^(-2x) + e^(-2x)ln(1+eˣ) + e^(-x)ln(1+eˣ) y(x) = c₁e^(-2x) + c₂e^(-x) - e^(-3x) + e^(-2x) ln(1+eˣ) - e^(-2x) + e^(-x) ln(1+eˣ) y(x) = c₁e^(-2x) + c₂e^(-x) - e^(-3x) + (e^(-2x) + e^(-x))ln(1 + eˣ) - e^(-2x) Question 4 Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. cos x dy/dx + (sin x)y = 1 Answer cos x dy/dx + sin x y = 1 By dividing both sides cos x dy/dx + (sin x / cos x) y = 1/cos x dy/dx + (tan x) y = sec x This is the form of the differential equation. dy/dx + P(x)y = f(x) With P(x) = tan x and f(x) = sec x P(x) is continuous on ( -π/2, π/2 ) Find the integrating factor I(x) I(x) = e^(∫ tan x dx) = e^( -ln|cos x|) = 1/|cos x| = 1/cos x Add the integrating factor to the differential equation. (1/cos x)dy/dx + (sin x/cos² x) y = sec²x (1/cos x) dy/dx + (sin x/cos x) * y / cosx = sec²x Write the left side in the form d/dx [I(x)y] d/dx[y/cos x] = sec²x d(y/cos x) = sec²x dx integrate both sides y/cos x = ∫ sec²x dx y/cos x = tan x + C Solve for y y = cos x tan x + C cos x y = sin x + C cos x Question 5 Use a linear approximation (or differentials) to estimate the given number. (1.999)⁴ Answer Step 1 We know that lim(∆x→0) (f(x + ∆x) - f(x)) / ∆x = f'(x) Hence for small values of ∆x; we can say that (f(x + ∆x) - f(x)) / ∆x ≈ f'(x) f(x + ∆x) ≈ f(x) + ∆x * f'(x) Let f(x) = x⁴, ∆x = -0.001, x = 2 Then we have f(2 -0.001) ≈ f(2) + (-0.001) * f'(2) (1.999)⁴ ≈ 2⁴ + (-0.001) * f'(2) → (1) The only Unknown we have is f'(2) Step 2 Finding f'(2) f(x) = x⁴ Differentiate f'(x) = 4x³ f'(2) = 4(2)³ = 32 Step 3 Substitute the value of f'(2) in Enq(1) (1.999)⁴ ≈ 2⁴ + (-0.001). (32) (1.999)⁴ ≈ 16 - 0.032 = 15.968 Result (1.999)⁴ ≈ 15.968