Questions & explanations
1. The solution of dy/dx = y^2 - 1 with y(0) = 0 is:
- y = (1 - e^{2x})/(1 + e^{2x})
- y = (e^{2x} - 1)/(e^{2x} + 1)
- y = (1 - e^{-2x})/(1 + e^{-2x})
- y = (e^{-2x} - 1)/(e^{-2x} + 1)
Answer: y = (e^{-2x} - 1)/(e^{-2x} + 1)
Separate variables: dy/(y^2-1) = dx. Use partial fractions: 1/(y^2-1) = 1/2[1/(y-1) - 1/(y+1)]. Integrate: (1/2) ln|(y-1)/(y+1)| = x + C. With y(0)=0, C=0. Since y(0)=0 lies in (-1,1), |(y-1)/(y+1)| = (1-y)/(1+y). Thus (1-y)/(1+y) = e^{2x}. Solve: y = (1 - e^{2x})/(1 + e^{2x}). Multiply numerator and denominator by e^{-2x}: y = (e^{-2x} - 1)/(e^{-2x} + 1).
2. The solution of (1 + y^2) dx - (tan^{-1} y - x) dy = 0 is:
- x = tan^{-1} y - 1 + C e^{tan^{-1} y}
- x = tan^{-1} y + 1 + C e^{-tan^{-1} y}
- x = tan^{-1} y - 1 + C e^{-tan^{-1} y}
- x = tan^{-1} y + 1 + C e^{tan^{-1} y}
Answer: x = tan^{-1} y - 1 + C e^{-tan^{-1} y}
Rewrite as dx/dy + x/(1+y^2) = tan^{-1} y/(1+y^2). This is linear in x with P(y)=1/(1+y^2), IF = e^{tan^{-1} y}. Then d/dy(x e^{tan^{-1} y}) = e^{tan^{-1} y} tan^{-1} y/(1+y^2). Substitute u=tan^{-1} y, integrate ∫ u e^u du = (u-1)e^u + C. Hence x e^{tan^{-1} y} = (tan^{-1} y - 1)e^{tan^{-1} y} + C, so x = tan^{-1} y - 1 + C e^{-tan^{-1} y}.
3. The solution of (1 + y^2) dx = (tan^{-1} y - x) dy is:
- x = tan^{-1} y + 1 + C e^{tan^{-1} y}
- x = tan^{-1} y + 1 + C e^{-tan^{-1} y}
- x = tan^{-1} y - 1 + C e^{tan^{-1} y}
- x = tan^{-1} y - 1 + C e^{-tan^{-1} y}
Answer: x = tan^{-1} y - 1 + C e^{-tan^{-1} y}
Rewrite as dx/dy + x/(1+y^2) = tan^{-1}y/(1+y^2). Integrating factor = e^{∫ dy/(1+y^2)} = e^{tan^{-1}y}. Multiply: d/dy(x e^{tan^{-1}y}) = tan^{-1}y e^{tan^{-1}y}/(1+y^2). Substitute u=tan^{-1}y, integrate ∫ u e^u du = (u-1)e^u + C. So x e^{tan^{-1}y} = (tan^{-1}y - 1)e^{tan^{-1}y} + C, giving x = tan^{-1}y - 1 + C e^{-tan^{-1}y}.
4. Solve the initial value problem: (x^2 + xy) dy = (x^2 + y^2) dx, y(1) = 0. Which of the following is the implicit solution?
- (x + y)^2 = x e^{y/x}
- (x - y)^2 = x e^{y/x}
- (x - y)^2 = x e^{-y/x}
- (x + y)^2 = x e^{-y/x}
Answer: (x - y)^2 = x e^{-y/x}
The DE is homogeneous. Substitute y=vx, dy=v dx+x dv. Simplify to x(1+v)dv = (1-v)dx. Separate: (1+v)/(1-v) dv = dx/x. Integrate: -ln|1-v| - v = ln|x|+C. Using y(1)=0 gives C=0. Then e^{-v}/(1-v)=x. Substitute v=y/x: e^{-y/x}/(1-y/x)=x => (x-y)/x = x e^{-y/x} => (x-y)^2 = x e^{-y/x}.
5. The solution of dy/dx + y cot x = 4x cosec x with y(π/2) = 0 is:
- y = (2x^2 - π^2/2) / sin x
- y = (2x^2 + π^2/2) / sin x
- y = (2x^2 - π^2/2) / cos x
- y = (2x^2 + π^2/2) / cos x
Answer: y = (2x^2 - π^2/2) / sin x
The equation is linear with P = cot x, Q = 4x cosec x. Integrating factor IF = e^{∫ cot x dx} = sin x. Multiplying gives d/dx(y sin x) = 4x. Integrate: y sin x = 2x^2 + C. Using y(π/2)=0 gives C = -π^2/2. Hence y = (2x^2 - π^2/2)/sin x.
6. Solve x dy/dx + y = y^2 log x. Find the general solution.
- 1/y = log x + 1 + C/x
- 1/y = log x + Cx
- 1/y = log x + 1 + C
- 1/y = log x + 1 + Cx
Answer: 1/y = log x + 1 + Cx
Divide by x: dy/dx + y/x = (log x/x) y^2, Bernoulli n=2. Substitute v=1/y: dv/dx - v/x = -log x/x. Linear in v, IF=1/x. d/dx(v/x) = -log x/x^2. Integrate: v/x = (log x)/x + 1/x + C, so v = log x + 1 + Cx. Hence 1/y = log x + 1 + Cx.
7. Which of the following differential equations is in variable separable form?
- dy/dx = x + y
- dy/dx = sin(x + y)
- dy/dx = x^2 + y^2
- dy/dx = x y
Answer: dy/dx = x y
A differential equation is variable separable if it can be written as dy/dx = f(x) g(y). Here, dy/dx = x y = (x)(y), so f(x)=x and g(y)=y. The other options cannot be factored into a product of a function of x and a function of y.
8. Which of the following is an ordinary differential equation?
- ∂u/∂t = k ∂²u/∂x²
- ∂u/∂x + ∂u/∂y = 0
- ∂²u/∂x² + ∂²u/∂y² = 0
- d²y/dx² + y = 0
Answer: d²y/dx² + y = 0
An ordinary differential equation involves derivatives with respect to only one independent variable. Option a uses ordinary derivatives d/dx, so it is an ODE. The other options contain partial derivatives (∂), making them PDEs.
9. A tank contains 100 L of pure water. Brine with 0.5 kg salt per L enters at 2 L/min. The well-stirred mixture leaves at 2 L/min. The amount of salt (in kg) in the tank after 50 minutes is:
- 50(1 - e^{-2})
- 50(1 - e^{-1/2})
- 50(1 - e^{-1})
- 50(1 - e^{-1/50})
Answer: 50(1 - e^{-1})
Using principle of mass balance: dS/dt = inflow rate - outflow rate = 2*0.5 - (S/100)*2 = 1 - S/50. Solve linear ODE: dS/dt + S/50 = 1, integrating factor e^{t/50}. Solution: S(t) = 50(1 - e^{-t/50}). At t=50, S=50(1-e^{-1}).
10. The general solution of x dy - y dx = sqrt(x^2 + y^2) dx is:
- y + sqrt(x^2 + y^2) = C
- y + sqrt(x^2 + y^2) = C x^2
- y - sqrt(x^2 + y^2) = C x^2
- y + sqrt(x^2 + y^2) = C x
Answer: y + sqrt(x^2 + y^2) = C x
The equation is homogeneous. Write dy/dx = y/x + sqrt(1+(y/x)^2). Substitute y=vx, get x dv/dx = sqrt(1+v^2). Separate: dv/sqrt(1+v^2) = dx/x. Integrate: sinh^{-1}(v) = ln|x| + C → v+sqrt(1+v^2) = Cx → y+sqrt(x^2+y^2) = Cx.
11. The general solution of x dy/dx + y = x^2 y^3 is:
- 1/(x^2 y^2) = C + 2 ln|x|
- 1/(x^2 y^2) = C - 2 ln|x|
- 1/(x y^2) = C - 2 ln|x|
- 1/(x^2 y) = C - 2 ln|x|
Answer: 1/(x^2 y^2) = C - 2 ln|x|
Divide by x: dy/dx + y/x = x y^3 (Bernoulli, n=3). Substitute v = y^{-2}, then dv/dx - 2v/x = -2x. IF = 1/x^2, so d/dx(v/x^2) = -2/x. Integrate: v/x^2 = -2 ln|x| + C. Since v = 1/y^2, we get 1/(x^2 y^2) = C - 2 ln|x|.
12. The differential equation of the family y = e^x (A cos x + B sin x) is:
- d²y/dx² + 2 dy/dx - 2y = 0
- d²y/dx² + 2 dy/dx + 2y = 0
- d²y/dx² - 2 dy/dx - 2y = 0
- d²y/dx² - 2 dy/dx + 2y = 0
Answer: d²y/dx² - 2 dy/dx + 2y = 0
Differentiate twice and eliminate A and B. First derivative: dy/dx = y + e^x(-A sin x + B cos x). Second derivative: d²y/dx² = dy/dx + (dy/dx - y) - y = 2 dy/dx - 2y. Rearranging gives d²y/dx² - 2 dy/dx + 2y = 0.