3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Consider the family of all circles in the xy-plane that pass through both points (1,0) and (−1,0). The differential equation of this family (with y′=dxdy) is ______. Exactly one option is correct.
A(x2−y2+1)y′=2xy
B(x2+y2−1)y′=2xy
C(x2−y2−1)y′=2xy
D(y2−x2−1)y′=2xy
Show answer and solution
Answer: C
A circle x2+y2+2gx+2fy+c=0 passes through (±1,0) iff 1+2g+c=0 and 1−2g+c=0, i.e. g=0, c=−1. So the family is x2+y2+2fy−1=0, one arbitrary constant f, hence a first-order equation.
Differentiate: 2x+2yy′+2fy′=0. From the family, 2f=y1−x2−y2 (on the circles, away from y=0).
Substituting: 2x+2yy′+y(1−x2−y2)y′=0. Multiply by y: 2xy+(2y2+1−x2−y2)y′=0, i.e. (y2−x2+1)y′+2xy=0.
Equivalently (x2−y2−1)y′=2xy.
Check: on x2+y2+2fy=1 we have y′=−y+fx; substituting this slope into the equation and using the circle equation gives 0 identically (verified symbolically). Each other option differs from it by a nonzero multiple of y′, so fails.
Trap: Sign slips when eliminating f; keep 2f=(1−x2−y2)/y exactly.
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
For x>0, let y=y(x) be the solution of xdxdy=y+xe−y/x with y(1)=0, and let I be the largest open interval containing 1 on which this solution exists. The pair (I, y(e)) is ______. Exactly one option is correct.
AI=(0,∞), y(e)=eln2
BI=(e1,∞), y(e)=eln2
CI=(e1,∞), y(e)=ln2
DI=(0,∞), y(e)=ln2
Show answer and solution
Answer: B
Put y=vx: x(v+xv′)=vx+xe−v, so xv′=e−v. Separate (no division by zero is involved since e−v>0): evdv=xdx, so ev=lnx+C.
y(1)=0 gives v(1)=0, C=1: ey/x=1+lnx, i.e. y=xln(1+lnx).
This needs 1+lnx>0, i.e. x>e1. As x→e1+, y→−∞, so the solution cannot be continued to the left; to the right it exists for all x. Hence I=(e1,∞).
Substitution check: y′=ln(1+lnx)+1+lnx1, so xy′=y+1+lnxx=y+xe−y/x; and y(1)=0.
y(e)=eln2.
Trap: Taking I=(0,∞) from the domain of the equation; the logarithm of 1+lnx ends the solution at x=1/e.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
A curve in the region y>0 passes through (e,1) and satisfies ydx=(x+y2ey)dy at each of its points. The ordinate of the point of the curve whose abscissa is 2e2 is ______. Exactly one option is correct.
Show answer and solution
Answer: D
The equation is not linear in y, but it is linear in x as a function of y: dydx−yx=yey.
Integrating factor e−∫dy/y=y1 (for y>0): dyd(yx)=ey, so yx=ey+C.
At (e,1): e=e+C, so C=0 and the curve is x=yey.
Check: dydx=(1+y)ey and y(1+y)ey=yey+y2ey=x+y2ey; and x(1)=e.
For y>0, x=yey is strictly increasing, so yey=2e2 has the single solution y=2.
Insight: Reading the equation as linear in x (with y the independent variable) is the key recognition step.
Difficulty: Very hard