First order differential equations practice questions.

Determine the integrating factor for the linear differential equation dy/dx + (2/x)y = x, where x > 0.

• e^(2x)
• x²
• 2 ln x
• 1/x²

Answer: x²

The integrating factor is e^(∫P dx) = e^(∫2/x dx) = e^(2 ln x) = x². Forgetting to simplify e^(2 ln x), or exponentiating incorrectly, are the usual slips.

Euler's method with step length h = 0.1 is applied to dy/dx = x + y with y(0) = 1. Calculate the approximation to y(0.2).

• 1.20
• 1.10
• 1.22
• 1.24

Answer: 1.22

Step 1: y₁ = 1 + 0.1(0 + 1) = 1.1 at x = 0.1. Step 2: y₂ = 1.1 + 0.1(0.1 + 1.1) = 1.22. Taking a single step of 0.2 gives the distractor 1.20.

A homogeneous differential equation has the form dy/dx = f(y/x). State the substitution that converts it into a separable equation.

• v = y + x
• x = vy
• v = xy
• y = vx

Answer: y = vx

Setting y = vx gives dy/dx = v + x·dv/dx, and since y/x = v the equation becomes v + x·dv/dx = f(v), which is separable in v and x.

Determine the general solution of the differential equation dy/dx = y/x for x > 0.

• y = Ax
• y = Ae^x
• y = x + A
• y² = x² + A

Answer: y = Ax

Separating: ∫(1/y) dy = ∫(1/x) dx gives ln|y| = ln x + c, so y = Ax. Students often wrongly produce an exponential because they expect e to appear whenever ln does.

Solve dy/dx + y = e^x given that y(0) = 1.

• y = e^x
• y = (e^x − e^(−x))/2
• y = (e^x + e^(−x))/2
• y = e^x/2 + 1/2

Answer: y = (e^x + e^(−x))/2

The integrating factor is e^x, so (e^x·y)' = e^(2x), giving e^x·y = e^(2x)/2 + C. With y(0) = 1, C = 1/2, so y = (e^x + e^(−x))/2.

The substitution y = vx is applied to the homogeneous equation dy/dx = (x² + y²)/(xy). Determine the resulting equation in v and x.

• x·dv/dx = v
• x·dv/dx = 1/v
• x·dv/dx = 1 + v²
• x·dv/dx = (1 + v²)/v

Answer: x·dv/dx = 1/v

With y = vx: v + x·dv/dx = (1 + v²)/v, so x·dv/dx = (1 + v²)/v − v = 1/v. Forgetting to subtract v gives the distractor (1 + v²)/v.