Question 1

The differential equation dy/dx = 2xy satisfies the initial condition y(0) = 1. What is y(x)?

Accepted Answer

This is a separable differential equation. We separate variables: dy/y = 2x dx. Integrating both sides: ∫(1/y) dy = ∫2x dx gives ln|y| = x² + C. Using the initial condition y(0) = 1: ln(1) = 0 + C, so C = 0. Therefore ln|y| = x², which gives y = e^(x²).

Question 2

The differential equation dy/dx = 2xy with initial condition y(0) = 1 has the solution:

Accepted Answer

This is a separable differential equation. We separate variables: dy/y = 2x dx. Integrating both sides: ∫(1/y)dy = ∫2x dx gives ln|y| = x² + C. Using the initial condition y(0) = 1: ln(1) = 0 + C, so C = 0. Therefore ln|y| = x², which gives y = e^(x²).

Question 3

The order and degree of the differential equation (d²y/dx²)² + (dy/dx)³ = x are respectively:

Accepted Answer

Order is determined by the highest derivative present. Here, d²y/dx² is the highest derivative, so order = 2. Degree is the power of the highest-order derivative when the equation is polynomial in derivatives. The highest-order derivative is d²y/dx², and it appears with power 2 in the term (d²y/dx²)². Therefore, degree = 2.

Question 4

The solution to the differential equation dy/dx + y = e^(-x) is:

Accepted Answer

This is a first-order linear differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = 1 and Q(x) = e^(-x). The integrating factor is μ(x) = e^(∫P(x)dx) = e^(∫1 dx) = e^x. Multiply both sides by e^x: e^x(dy/dx) + e^x·y = e^x·e^(-x) = 1. The left side is d/dx(e^x·y). Integrating: e^x·y = ∫1 dx = x + C. Therefore, y = (x + C)e^(-x).

Question 5

If y = e^(2x)·sin(3x), then d²y/dx² equals:

Accepted Answer

We use the product rule twice. First derivative: dy/dx = d/dx(e^(2x))·sin(3x) + e^(2x)·d/dx(sin(3x)) = 2e^(2x)·sin(3x) + e^(2x)·3cos(3x) = e^(2x)[2sin(3x) + 3cos(3x)]. Second derivative: d²y/dx² = d/dx{e^(2x)[2sin(3x) + 3cos(3x)]}. Using product rule: d²y/dx² = 2e^(2x)[2sin(3x) + 3cos(3x)] + e^(2x)[6cos(3x) - 9sin(3x)] = e^(2x)[4sin(3x) + 6cos(3x) + 6cos(3x) - 9sin(3x)] = e^(2x)[-5sin(3x) + 12cos(3x)].

Question 6

The differential equation (1 + x²)dy/dx + 2xy = 4x² is linear. Its solution is:

Accepted Answer

Rewrite in standard form by dividing by (1 + x²): dy/dx + [2x/(1 + x²)]y = 4x²/(1 + x²). Here P(x) = 2x/(1 + x²) and Q(x) = 4x²/(1 + x²). The integrating factor is μ(x) = e^(∫[2x/(1+x²)]dx). Note that ∫[2x/(1+x²)]dx = ln(1 + x²) (by substitution u = 1 + x²). So μ(x) = e^(ln(1+x²)) = 1 + x². Multiply the equation by (1 + x²): (1 + x²)dy/dx + 2xy = 4x². The left side is d/dx[(1 + x²)y]. Integrating: (1 + x²)y = ∫4x² dx = (4x³/3) + C. Therefore, y = (4x³/3 + C)/(1 + x²).

Question 7

The order and degree of the differential equation d²y/dx² + (dy/dx)³ = sin(x) are respectively:

Accepted Answer

Order of a differential equation is the highest derivative present. Here, the highest derivative is d²y/dx² (second derivative), so order = 2. Degree is the power of the highest-order derivative when the equation is expressed as a polynomial in derivatives. The highest-order derivative d²y/dx² appears with power 1, so degree = 1. Therefore, order = 2 and degree = 1.

Question 8

If y = e^(2x)·sin(3x), then dy/dx equals:

Accepted Answer

This requires the product rule: d/dx[u·v] = u'v + uv', where u = e^(2x) and v = sin(3x). First, u' = 2e^(2x) (by chain rule). Second, v' = 3cos(3x) (by chain rule). Therefore, dy/dx = 2e^(2x)·sin(3x) + e^(2x)·3cos(3x) = e^(2x)[2sin(3x) + 3cos(3x)].

Question 9

The solution of the differential equation (1 + x²)dy/dx = xy with y(0) = 2 is:

Accepted Answer

This is a separable differential equation. Separate variables: dy/y = x·dx/(1 + x²). Integrate both sides: ∫(1/y) dy = ∫x/(1 + x²) dx. The left side gives ln|y|. For the right side, use substitution u = 1 + x², du = 2x dx, so ∫x/(1 + x²) dx = (1/2)∫(1/u) du = (1/2)ln|u| = (1/2)ln(1 + x²). Therefore, ln|y| = (1/2)ln(1 + x²) + C. Apply y(0) = 2: ln(2) = (1/2)ln(1) + C = 0 + C, so C = ln(2). Thus ln|y| = (1/2)ln(1 + x²) + ln(2), which gives y = 2√(1 + x²).

Question 10

The differential equation d²y/dx² + 4y = 0 has the general solution y = A·cos(2x) + B·sin(2x). If y(0) = 3 and dy/dx|_(x=0) = 4, what is the value of A + B?

Accepted Answer

Given y = A·cos(2x) + B·sin(2x). From y(0) = 3: A·cos(0) + B·sin(0) = 3, so A = 3. Now differentiate: dy/dx = -2A·sin(2x) + 2B·cos(2x). From dy/dx|_(x=0) = 4: -2A·sin(0) + 2B·cos(0) = 4, so 2B = 4, giving B = 2. Therefore A + B = 3 + 2 = 5.

Question 11

Consider the differential equation dy/dx = (y² + 1)/x. Which of the following is the correct form after separating variables?

Accepted Answer

The equation dy/dx = (y² + 1)/x is separable. Rearranging: dy/(y² + 1) = dx/x. This is the correct separated form, ready for integration on both sides.

Question 12

The differential equation d²y/dx² - 5·dy/dx + 6y = 0 is a second-order linear homogeneous equation. What is the characteristic equation?

Accepted Answer

For a second-order linear homogeneous differential equation of the form d²y/dx² + p·dy/dx + q·y = 0, the characteristic equation is obtained by substituting y = e^(rx), which gives r² + p·r + q = 0. For our equation d²y/dx² - 5·dy/dx + 6y = 0, the characteristic equation is r² - 5r + 6 = 0.

Question 13

The solution to the differential equation dy/dx = 3e^(2x) with y(0) = 5 is:

Accepted Answer

This is a direct integration problem. Integrate both sides: ∫dy = ∫3e^(2x) dx. The left side gives y. For the right side, use the formula ∫e^(ax) dx = (1/a)e^(ax): ∫3e^(2x) dx = 3·(1/2)e^(2x) = (3/2)e^(2x) + C. So y = (3/2)e^(2x) + C. Apply y(0) = 5: 5 = (3/2)e^0 + C = 3/2 + C, so C = 5 - 3/2 = 7/2. Therefore y = (3/2)e^(2x) + 7/2.

Question 14

The differential equation dy/dx = (y - x)/(y + x) can be solved by the substitution y = vx. After substitution and simplification, the resulting separable equation is:

Accepted Answer

Given: dy/dx = (y - x)/(y + x)

Step 1: Use substitution y = vx, where v is a function of x.

Step 2: Differentiate y = vx:
dy/dx = v + x(dv/dx)

Step 3: Substitute into the original equation:
v + x(dv/dx) = (vx - x)/(vx + x)
v + x(dv/dx) = x(v - 1)/(x(v + 1))
v + x(dv/dx) = (v - 1)/(v + 1)

Step 4: Rearrange:
x(dv/dx) = (v - 1)/(v + 1) - v

Step 5: Simplify the right side:
x(dv/dx) = (v - 1)/(v + 1) - v(v + 1)/(v + 1)
x(dv/dx) = [(v - 1) - v(v + 1)]/(v + 1)
x(dv/dx) = [v - 1 - v² - v]/(v + 1)
x(dv/dx) = [-v² - 1]/(v + 1)
x(dv/dx) = -(v² + 1)/(v + 1)

Step 6: Separate variables:
(v + 1)dv/(v² + 1) = -dx/x

This is now separable.

Question 15

The differential equation dy/dx = (y² + 2xy)/(x²) can be reduced to a linear form by the substitution v = y/x. After substitution, the resulting differential equation in v and x is:

Accepted Answer

Given: dy/dx = (y² + 2xy)/x²

Step 1: Recognize this is a homogeneous differential equation. Use substitution v = y/x, so y = vx.

Step 2: Differentiate y = vx with respect to x:
dy/dx = v + x(dv/dx)

Step 3: Substitute into the original equation:
v + x(dv/dx) = (v²x² + 2x·vx)/x²
v + x(dv/dx) = (v²x² + 2vx²)/x²
v + x(dv/dx) = v² + 2v

Step 4: Simplify:
x(dv/dx) = v² + 2v - v
x(dv/dx) = v² + v

Step 5: Separate variables:
dv/(v² + v) = dx/x

This is now a separable form. The linear form (after partial fractions) is:
dv/[v(v+1)] = dx/x

Which gives: (1/v - 1/(v+1))dv = dx/x

Therefore, the correct reduced form is: x(dv/dx) = v(v + 1) or equivalently dv/(v² + v) = dx/x

Question 16

The solution to the differential equation dy/dx + y·cot(x) = 2x·csc(x) with the condition y(π/2) = 1 is:

Accepted Answer

This is a first-order linear differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = cot(x) and Q(x) = 2x·csc(x). The integrating factor is μ(x) = e^(∫cot(x)dx) = e^(ln|sin(x)|) = sin(x). Multiply both sides by sin(x): sin(x)·dy/dx + sin(x)·cot(x)·y = 2x. Note that sin(x)·cot(x) = cos(x), so: sin(x)·dy/dx + cos(x)·y = 2x. The LHS is d/dx[y·sin(x)], so: d/dx[y·sin(x)] = 2x. Integrate both sides: y·sin(x) = ∫2x dx = x² + C. Therefore: y = (x² + C)/sin(x). Apply the initial condition y(π/2) = 1: 1 = (π²/4 + C)/sin(π/2) = π²/4 + C. Thus C = 1 - π²/4. The solution is y = (x² + 1 - π²/4)/sin(x) = [x² - (π²/4 - 1)]/sin(x). Simplifying: y = (x² + 1 - π²/4)/sin(x) or equivalently y·sin(x) = x² + 1 - π²/4.

Question 17

Consider the differential equation (1 + x²)dy/dx + 2xy = 4x². The general solution is:

Accepted Answer

This is a first-order linear differential equation. Rewrite in standard form: dy/dx + (2xy)/(1+x²) = 4x²/(1+x²). Here P(x) = 2x/(1+x²) and Q(x) = 4x²/(1+x²). The integrating factor is μ(x) = e^(∫2x/(1+x²)dx). Let u = 1+x², then du = 2x dx, so ∫2x/(1+x²)dx = ln(1+x²). Thus μ(x) = e^(ln(1+x²)) = 1+x². Multiply the equation by (1+x²): (1+x²)dy/dx + 2xy = 4x². The LHS is d/dx[y(1+x²)], so: d/dx[y(1+x²)] = 4x². Integrate both sides: y(1+x²) = ∫4x² dx = (4x³)/3 + C. Therefore: y = (4x³/3 + C)/(1+x²) = (4x³ + 3C)/(3(1+x²)).

Question 18

The differential equation d²y/dx² - 5dy/dx + 6y = 0 has the general solution:

Accepted Answer

This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is obtained by substituting y = e^(mx): m² - 5m + 6 = 0. Factoring: (m - 2)(m - 3) = 0, so m = 2 or m = 3. Since we have two distinct real roots, the general solution is y = C₁e^(2x) + C₂e^(3x), where C₁ and C₂ are arbitrary constants.

Question 19

The solution to the differential equation dy/dx = (y - x)/(y + x) that passes through the point (1, 2) is:

Accepted Answer

This is a homogeneous differential equation. Use the substitution v = y/x, so y = vx and dy/dx = v + x(dv/dx). Substitute into the equation: v + x(dv/dx) = (vx - x)/(vx + x) = (v - 1)/(v + 1). Rearranging: x(dv/dx) = (v - 1)/(v + 1) - v = (v - 1 - v(v + 1))/(v + 1) = (v - 1 - v² - v)/(v + 1) = (-1 - v²)/(v + 1). Thus: x(dv/dx) = -(1 + v²)/(v + 1). Separating variables: (v + 1)/(1 + v²) dv = -dx/x. Integrate both sides: ∫(v + 1)/(1 + v²) dv = -∫dx/x. Split the LHS: ∫v/(1+v²) dv + ∫1/(1+v²) dv = (1/2)ln(1+v²) + arctan(v). So: (1/2)ln(1+v²) + arctan(v) = -ln|x| + C. Substitute v = y/x: (1/2)ln(1 + y²/x²) + arctan(y/x) = -ln|x| + C. Simplify: (1/2)ln((x² + y²)/x²) + arctan(y/x) = -ln|x| + C, which gives (1/2)ln(x² + y²) - ln|x| + arctan(y/x) = -ln|x| + C. Thus: (1/2)ln(x² + y²) + arctan(y/x) = C. Apply the initial condition (1, 2): (1/2)ln(1 + 4) + arctan(2) = C, so C = (1/2)ln(5) + arctan(2). The solution is: (1/2)ln(x² + y²) + arctan(y/x) = (1/2)ln(5) + arctan(2), or equivalently: ln(x² + y²) + 2arctan(y/x) = ln(5) + 2arctan(2).

Question 20

The differential equation dy/dx = (y² + 2xy)/(x²) can be transformed into a linear differential equation by the substitution v = y/x. After substitution, the resulting differential equation in v and x is:

Accepted Answer

Given: dy/dx = (y² + 2xy)/x²

Step 1: Let v = y/x, so y = vx.

Step 2: Differentiate y = vx with respect to x using the product rule:
dy/dx = v + x(dv/dx)

Step 3: Substitute into the original equation:
v + x(dv/dx) = (v²x² + 2x·vx)/x²
v + x(dv/dx) = (v²x² + 2vx²)/x²
v + x(dv/dx) = v² + 2v

Step 4: Simplify:
x(dv/dx) = v² + 2v - v
x(dv/dx) = v² + v

Step 5: Rearrange into standard linear form:
dv/dx = (v² + v)/x
OR equivalently: x(dv/dx) - v² - v = 0

This is a separable form: dv/(v² + v) = dx/x, which can be solved by partial fractions.

The correct transformed equation is: dv/dx - v/x = v²/x

CAPF AC Differential Equations

Differential Equations— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Differential Equations — Practice Questions

60-Second Revision — Differential Equations