Module 6 · JEE Mathematics
Differential Equations — The Equations That Know the Future
dy/dx tells you the rate. The DE tells you the whole story. Population growth, cooling coffee, falling objects — it's all DEs.
📝 What is a DE?
🔄 Variable Sep.
🏠 Homogeneous
📏 Linear DE
🎯 Practice
📝 Part 1: What Is a Differential Equation? — The Equation That Knows the Future
🔄 Part 2: Variable Separable — Separate, Integrate, Done
🏠 Part 3: Homogeneous DE — When dy/dx = f(y/x)
📏 Part 4: Linear First-Order DE — The Integrating Factor
⚠️ Common Mistakes (Don't Fall For These!)
🧠 Differential Equations Toolkit
🔥 Advanced Worked Examples (JEE Level)
🎯 Practice Zone: Test Your Understanding
A Differential Equation Describes HOW Something Changes
A differential equation is an equation that involves a function and its derivatives. It describes a relationship between a quantity and its rate of change.
📝 The Core Idea
DE = "I Know How Fast It's Changing — Tell Me What It Is"
Regular equation: x² + 3 = 7 → solve for x. One answer.
Differential equation: dy/dx = 2x → solve for y(x). A whole family of curves!
Why "knows the future"? If you know the rate of change of a population (dP/dt = kP), and you know the current population, you can predict the population at ANY future time! That's the power of DEs — they model how things evolve.
Examples of DEs in nature:
• Radioactive decay: dN/dt = −λN (rate of decay ∝ amount left)
• Newton's cooling: dT/dt = −k(T − T₀) (cooling rate ∝ temperature difference)
• Population growth: dP/dt = rP (growth rate ∝ current population)
Differential equation: dy/dx = 2x → solve for y(x). A whole family of curves!
Why "knows the future"? If you know the rate of change of a population (dP/dt = kP), and you know the current population, you can predict the population at ANY future time! That's the power of DEs — they model how things evolve.
Examples of DEs in nature:
• Radioactive decay: dN/dt = −λN (rate of decay ∝ amount left)
• Newton's cooling: dT/dt = −k(T − T₀) (cooling rate ∝ temperature difference)
• Population growth: dP/dt = rP (growth rate ∝ current population)
💡 dy/dx = rate of change. A DE = an equation relating y and dy/dx. Solving it = finding y(x).
Order and Degree — The DE's ID Card
Order = the highest derivative in the equation. Degree = the power of that highest derivative (after removing radicals and fractions).
📊 Order & Degree — Classifying DEs
Order = how many times you differentiate (the number of dashes). Degree = the exponent on the highest derivative. If the DE has sin(dy/dx), e^(dy/dx), or √(dy/dx), its degree is not defined (non-polynomial in derivatives).
General Solution vs Particular Solution
📝 Solutions
General Solution = Family of Curves. Particular Solution = One Specific Curve.
General solution: Contains arbitrary constants (one constant for each order). dy/dx = 2x → y = x² + C. The constant C represents a whole family of parabolas.
Particular solution: The constant(s) are determined by initial conditions (extra info like y(0) = 3). So C = 3, giving y = x² + 3.
Order determines constants: A first-order DE needs 1 initial condition. A second-order DE needs 2.
Particular solution: The constant(s) are determined by initial conditions (extra info like y(0) = 3). So C = 3, giving y = x² + 3.
Order determines constants: A first-order DE needs 1 initial condition. A second-order DE needs 2.
💡 Order = number of constants in the general solution = number of initial conditions needed for a particular solution.
The Simpliest DE: dy/dx = f(x)·g(y) — Just Separate and Integrate
If you can rearrange the DE so that all x's are on one side and all y's on the other, you're done. Just separate and integrate both sides.
🔄 Variable Separable
Form: dy/dx = f(x)·g(y) → ∫ 1/g(y) dy = ∫ f(x) dx
Steps:
1. Rearrange to get all y terms with dy and all x terms with dx.
2. Integrate both sides.
3. Solve for y if possible.
Example: dy/dx = 2xy.
Separate: dy/y = 2x dx.
Integrate: ∫ dy/y = ∫ 2x dx → ln|y| = x² + C.
Solve: |y| = e^(x²+C) = e^C·e^(x²). Let A = ±e^C.
1. Rearrange to get all y terms with dy and all x terms with dx.
2. Integrate both sides.
3. Solve for y if possible.
Example: dy/dx = 2xy.
Separate: dy/y = 2x dx.
Integrate: ∫ dy/y = ∫ 2x dx → ln|y| = x² + C.
Solve: |y| = e^(x²+C) = e^C·e^(x²). Let A = ±e^C.
y = A·e^(x²)
Check: dy/dx = A·2x·e^(x²) = 2xy ✅
💡 "Separate and integrate" — if you can split x's and y's, the DE is solved. This is the most common JEE Main DE type!
Worked Example: Variable Separable
Solve dy/dx = 1 + y² given y(0) = 1. Find y(π/4).
1
Separate and integrate
▶
dy/dx = 1 + y²
dy/(1 + y²) = dx
∫ dy/(1 + y²) = ∫ dx
tan⁻¹(y) = x + C
dy/(1 + y²) = dx
∫ dy/(1 + y²) = ∫ dx
tan⁻¹(y) = x + C
2
Use the initial condition and evaluate
▶
y(0) = 1: tan⁻¹(1) = 0 + C → π/4 = C.
So tan⁻¹(y) = x + π/4.
y = tan(x + π/4).
At x = π/4: y = tan(π/4 + π/4) = tan(π/2) = undefined (infinite).
This means the solution blows up at x = π/4 — the function has a vertical asymptote there.
So tan⁻¹(y) = x + π/4.
y = tan(x + π/4).
At x = π/4: y = tan(π/4 + π/4) = tan(π/2) = undefined (infinite).
This means the solution blows up at x = π/4 — the function has a vertical asymptote there.
If Every Term Has the Same Total Degree, Substitute y = vx
A DE is homogeneous if dy/dx can be written as a function of y/x. The substitution y = vx (where v is a function of x) turns it into a variable-separable equation.
🏠 Homogeneous DE
Form: dy/dx = F(y/x). Substitute y = vx, then dy/dx = v + x·dv/dx
The substitution trick: Let y = vx (where v = y/x). Then:
dy/dx = v + x·dv/dx (product rule).
The DE becomes: v + x·dv/dx = F(v).
Rearranged: x·dv/dx = F(v) − v.
Separated: dv/(F(v) − v) = dx/x.
Now it's variable-separable! Integrate, then substitute back v = y/x.
How to check if a DE is homogeneous: Replace x with tx and y with ty. If all terms get the same power of t, it's homogeneous. Example: (x² + y²)dx + xy·dy = 0. Each term has degree 2.
dy/dx = v + x·dv/dx (product rule).
The DE becomes: v + x·dv/dx = F(v).
Rearranged: x·dv/dx = F(v) − v.
Separated: dv/(F(v) − v) = dx/x.
Now it's variable-separable! Integrate, then substitute back v = y/x.
How to check if a DE is homogeneous: Replace x with tx and y with ty. If all terms get the same power of t, it's homogeneous. Example: (x² + y²)dx + xy·dy = 0. Each term has degree 2.
💡 Homogeneous DE → substitute y = vx. The substitution converts it into a separable equation. v + x·dv/dx = F(v) is the key equation!
Worked Example: Homogeneous DE
Solve (x² − y²)dx + xy·dy = 0.
1
Rewrite in the form dy/dx = F(y/x)
▶
(x² − y²)dx + xy·dy = 0
xy·dy = (y² − x²)dx
dy/dx = (y² − x²)/(xy) = y/x − x/y.
This is dy/dx = y/x − x/y = F(y/x) where F(v) = v − 1/v. ✅ Homogeneous.
xy·dy = (y² − x²)dx
dy/dx = (y² − x²)/(xy) = y/x − x/y.
This is dy/dx = y/x − x/y = F(y/x) where F(v) = v − 1/v. ✅ Homogeneous.
2
Substitute y = vx
▶
y = vx, dy/dx = v + x·dv/dx.
v + x·dv/dx = v − 1/v
x·dv/dx = −1/v
v·dv = −dx/x
∫ v dv = −∫ dx/x
v²/2 = −ln|x| + C
v² = −2ln|x| + 2C = −ln(x²) + C₁
v + x·dv/dx = v − 1/v
x·dv/dx = −1/v
v·dv = −dx/x
∫ v dv = −∫ dx/x
v²/2 = −ln|x| + C
v² = −2ln|x| + 2C = −ln(x²) + C₁
3
Substitute back v = y/x
▶
(y/x)² = C₁ − ln(x²)
y² = x²(C₁ − ln(x²))
This is the general solution — a family of curves. The constant C₁ is determined by an initial condition.
Form: dy/dx + P(x)·y = Q(x). The Answer Is Always y·IF = ∫Q·IF dx
A linear first-order DE has the form dy/dx + P(x)·y = Q(x). The integrating factor (IF) method is the standard solution technique.
📏 Linear DE
dy/dx + P·y = Q → Multiply by IF = e^(∫P dx), then the LHS becomes d/dx(y·IF)
The magic: Multiply both sides by the integrating factor IF = e^(∫P dx). The left side becomes the derivative of (y·IF).
Solution formula:
1. Write the DE as dy/dx + Py = Q (P and Q are functions of x only).
2. Compute IF = e^(∫P dx).
3. Multiply both sides by IF: IF·dy/dx + IF·P·y = IF·Q.
4. Notice the LHS = d/dx(y·IF).
5. Integrate: y·IF = ∫ Q·IF dx + C.
6. Solve for y.
Example: dy/dx + 3y = eˣ. P = 3, Q = eˣ.
IF = e^(∫3 dx) = e^(3x).
y·e^(3x) = ∫ eˣ·e^(3x) dx = ∫ e^(4x) dx = e^(4x)/4 + C.
y = e^(4x)/(4e^(3x)) + C/e^(3x) = eˣ/4 + C·e^(−3x).
Solution formula:
y·IF = ∫ Q·IF dx + C
Steps:1. Write the DE as dy/dx + Py = Q (P and Q are functions of x only).
2. Compute IF = e^(∫P dx).
3. Multiply both sides by IF: IF·dy/dx + IF·P·y = IF·Q.
4. Notice the LHS = d/dx(y·IF).
5. Integrate: y·IF = ∫ Q·IF dx + C.
6. Solve for y.
Example: dy/dx + 3y = eˣ. P = 3, Q = eˣ.
IF = e^(∫3 dx) = e^(3x).
y·e^(3x) = ∫ eˣ·e^(3x) dx = ∫ e^(4x) dx = e^(4x)/4 + C.
y = e^(4x)/(4e^(3x)) + C/e^(3x) = eˣ/4 + C·e^(−3x).
💡 IF = e^(∫P dx) transforms the LHS into a perfect derivative. Then it's just integrate-and-multiply. This is the single most important method for first-order DEs.
Worked Example: Linear DE — Newton's Cooling
A cup of coffee at 90°C is placed in a room at 20°C. After 5 minutes, it's 60°C. Newton's law: dT/dt = −k(T − 20). Find when it reaches 40°C.
1
Set up and solve the DE
▶
dT/dt = −k(T − 20). Let's rewrite: dT/dt + kT = 20k.
This is linear with P = k, Q = 20k.
IF = e^(∫k dt) = e^(kt).
T·e^(kt) = ∫ 20k·e^(kt) dt = 20e^(kt) + C.
T = 20 + C·e^(−kt).
At t = 0, T = 90: 90 = 20 + C → C = 70.
So T(t) = 20 + 70·e^(−kt).
This is linear with P = k, Q = 20k.
IF = e^(∫k dt) = e^(kt).
T·e^(kt) = ∫ 20k·e^(kt) dt = 20e^(kt) + C.
T = 20 + C·e^(−kt).
At t = 0, T = 90: 90 = 20 + C → C = 70.
So T(t) = 20 + 70·e^(−kt).
2
Find k and solve for the target time
▶
At t = 5, T = 60: 60 = 20 + 70·e^(−5k)
40 = 70·e^(−5k)
e^(−5k) = 4/7 → −5k = ln(4/7) → k = −ln(4/7)/5 ≈ 0.1119.
Find t when T = 40: 40 = 20 + 70·e^(−kt)
20 = 70·e^(−kt) → e^(−kt) = 2/7
−kt = ln(2/7) → t = −ln(2/7)/k ≈ 11.2 minutes.
40 = 70·e^(−5k)
e^(−5k) = 4/7 → −5k = ln(4/7) → k = −ln(4/7)/5 ≈ 0.1119.
Find t when T = 40: 40 = 20 + 70·e^(−kt)
20 = 70·e^(−kt) → e^(−kt) = 2/7
−kt = ln(2/7) → t = −ln(2/7)/k ≈ 11.2 minutes.
The coffee reaches 40°C after about 11.2 minutes.
🌟 DE Solution Flowchart
Step 1: Can you separate? → Variable separable. Step 2: Is it dy/dx = F(y/x)? → Homogeneous (y = vx). Step 3: Is it dy/dx + Py = Q? → Linear (IF method).
If none of these, it may be Bernoulli (dy/dx + Py = Q·yⁿ) — transform into linear by letting v = y^(1−n).
These Traps Trip Up Even Top Students
⚠️ Mistake 1
Forgetting the + C When Integrating
Wrong: ∫ dy/y = ∫ 2x dx → ln|y| = x² → y = e^(x²) ❌
Right: ∫ dy/y = ∫ 2x dx → ln|y| = x² + C → y = ±e^C·e^(x²) = A·e^(x²) ✅
The constant of integration is ABSOLUTELY CRITICAL in DEs. Without it, you lose the entire family of solutions. The general solution MUST have constants equal to the order of the DE.
Right: ∫ dy/y = ∫ 2x dx → ln|y| = x² + C → y = ±e^C·e^(x²) = A·e^(x²) ✅
The constant of integration is ABSOLUTELY CRITICAL in DEs. Without it, you lose the entire family of solutions. The general solution MUST have constants equal to the order of the DE.
⚠️ Never omit + C when integrating a DE. The constant represents which specific curve you're talking about!
⚠️ Mistake 2
Confusing Degree and Order
Wrong: (d²y/dx²)³ + dy/dx = 0 → Order 3, Degree 2 ❌
Right: The highest derivative is d²y/dx² (order 2). The power of the highest derivative is 3. So Order = 2, Degree = 3 ✅.
Order always = the number of dashes on y (or the highest derivative). Degree = the exponent on that highest derivative (after removing radicals and fractions).
Right: The highest derivative is d²y/dx² (order 2). The power of the highest derivative is 3. So Order = 2, Degree = 3 ✅.
Order always = the number of dashes on y (or the highest derivative). Degree = the exponent on that highest derivative (after removing radicals and fractions).
⚠️ Order = highest derivative (count the dashes). Degree = exponent on that derivative (after cleaning radicals). Don't swap them!
⚠️ Mistake 3
Forgetting the Absolute Value in ln|y|
Wrong: ∫ dy/y = ∫ dx → ln(y) = x + C → y = e^(x+C) ❌
Right: ∫ dy/y = ∫ dx → ln|y| = x + C → |y| = e^(x+C) → y = ±e^C·e^x = A·e^x ✅
The absolute value matters because y could be negative. The constant A = ±e^C absorbs both the sign and the magnitude.
Right: ∫ dy/y = ∫ dx → ln|y| = x + C → |y| = e^(x+C) → y = ±e^C·e^x = A·e^x ✅
The absolute value matters because y could be negative. The constant A = ±e^C absorbs both the sign and the magnitude.
⚠️ ∫ dy/y = ln|y| + C, not ln(y) + C. The absolute value handles negative solutions correctly.
5 Strategies That Crack Almost Any Problem
🧠 Strategy 1
"Identify the Type First" — The DE Classification Flowchart
Before doing ANY algebra, look at the DE and ask:
1. Can I separate? → dy/dx = f(x)g(y)? → Separate and integrate.
2. Is it dy/dx = F(y/x)? → Homogeneous. Substitute y = vx.
3. Is it dy/dx + P(x)·y = Q(x)? → Linear. Find IF = e^(∫P dx).
4. Is it dy/dx + P·y = Q·yⁿ? → Bernoulli. Let v = y^(1−n) to make it linear.
5. None of the above? → Try an exact DE or look for a substitution.
Pro tip: Most JEE DE problems are designed to be one of these types. Identify the type first, then apply the standard method.
1. Can I separate? → dy/dx = f(x)g(y)? → Separate and integrate.
2. Is it dy/dx = F(y/x)? → Homogeneous. Substitute y = vx.
3. Is it dy/dx + P(x)·y = Q(x)? → Linear. Find IF = e^(∫P dx).
4. Is it dy/dx + P·y = Q·yⁿ? → Bernoulli. Let v = y^(1−n) to make it linear.
5. None of the above? → Try an exact DE or look for a substitution.
Pro tip: Most JEE DE problems are designed to be one of these types. Identify the type first, then apply the standard method.
💡 90% of JEE DE problems are one of: separable, homogeneous, or linear. Identify the type, apply the method, get the answer.
🧠 Strategy 2
"Check the Form Before Rearranging"
Don't blindly rearrange terms. First check if the DE is already in a standard form:
• dx/dy form: Sometimes the equation is easier if you treat x as a function of y. dx/dy + P(y)·x = Q(y) → linear in x!
• Symmetric form: M(x,y)dx + N(x,y)dy = 0. Check if it's homogeneous by looking at the degrees of M and N.
Example: (x² + y²)dx + xy·dy = 0. This is in symmetric form. The degree of each term is 2 → homogeneous.
• dx/dy form: Sometimes the equation is easier if you treat x as a function of y. dx/dy + P(y)·x = Q(y) → linear in x!
• Symmetric form: M(x,y)dx + N(x,y)dy = 0. Check if it's homogeneous by looking at the degrees of M and N.
Example: (x² + y²)dx + xy·dy = 0. This is in symmetric form. The degree of each term is 2 → homogeneous.
💡 Sometimes treating x as the dependent variable (dx/dy instead of dy/dx) makes a non-linear DE become linear. Always check both orientations!
🧠 Strategy 3
"Verify Your Solution by Substituting Back"
The fastest way to check your answer: plug the solution back into the original DE and verify it works.
If you got y = A·e^(x²) from dy/dx = 2xy:
dy/dx = 2x·A·e^(x²) = 2xy. ✅ Works!
This catches differentiation errors, sign errors, and integration constant mistakes. Always do this check in exams if you have time.
If you got y = A·e^(x²) from dy/dx = 2xy:
dy/dx = 2x·A·e^(x²) = 2xy. ✅ Works!
This catches differentiation errors, sign errors, and integration constant mistakes. Always do this check in exams if you have time.
💡 Substituting back is the ultimate proof. If your solution satisfies the original DE, it's correct. Period.
🧠 Strategy 4
"Don't Fear Initial Conditions — They Just Find the Constant"
Many students find DEs hard because they don't know what to do with y(0) = 5 or y'(0) = 2.
It's simple: First find the general solution (with constant C). Then plug in the initial condition to find C.
Example: dy/dx = 2xy, y(0) = 5.
General: y = A·e^(x²).
Plug x = 0, y = 5: 5 = A·e⁰ = A. So y = 5·e^(x²).
That's it! The initial condition just pinpoints which curve from the family you want.
It's simple: First find the general solution (with constant C). Then plug in the initial condition to find C.
Example: dy/dx = 2xy, y(0) = 5.
General: y = A·e^(x²).
Plug x = 0, y = 5: 5 = A·e⁰ = A. So y = 5·e^(x²).
That's it! The initial condition just pinpoints which curve from the family you want.
💡 Initial conditions come LAST. Solve the DE generally first, then use the initial conditions to find the constants.
Multi-Step Problems That Combine Multiple Concepts
Example 1: Bernoulli's Equation
Solve dy/dx + y/x = y².
1
Recognize it as Bernoulli: dy/dx + P·y = Q·yⁿ
▶
dy/dx + (1/x)·y = 1·y².
Here P = 1/x, Q = 1, n = 2.
Bernoulli's substitution: v = y^(1−n) = y^(1−2) = y^(−1) = 1/y.
Then dv/dx = −y^(−2)·dy/dx = −v²·dy/dx.
Here P = 1/x, Q = 1, n = 2.
Bernoulli's substitution: v = y^(1−n) = y^(1−2) = y^(−1) = 1/y.
Then dv/dx = −y^(−2)·dy/dx = −v²·dy/dx.
2
Transform to linear DE in v
▶
Divide the original DE by y²: y^(−2)·dy/dx + (1/x)·y^(−1) = 1.
Note: dv/dx = −y^(−2)·dy/dx, so y^(−2)·dy/dx = −dv/dx.
And y^(−1) = v.
So: −dv/dx + (1/x)·v = 1 → dv/dx − (1/x)·v = −1.
This is linear in v: dv/dx + P'·v = Q' where P' = −1/x, Q' = −1.
IF = e^(∫−1/x dx) = e^(−ln|x|) = 1/x.
v·(1/x) = ∫ −1·(1/x) dx = −ln|x| + C.
v = x(−ln|x| + C) = −x·ln|x| + Cx.
Note: dv/dx = −y^(−2)·dy/dx, so y^(−2)·dy/dx = −dv/dx.
And y^(−1) = v.
So: −dv/dx + (1/x)·v = 1 → dv/dx − (1/x)·v = −1.
This is linear in v: dv/dx + P'·v = Q' where P' = −1/x, Q' = −1.
IF = e^(∫−1/x dx) = e^(−ln|x|) = 1/x.
v·(1/x) = ∫ −1·(1/x) dx = −ln|x| + C.
v = x(−ln|x| + C) = −x·ln|x| + Cx.
3
Substitute back v = 1/y
▶
Since v = 1/y, the general solution is:
1/y = Cx − x·ln|x| → y = 1/(Cx − x·ln|x|)
This is the general solution. For JEE, the answer is often left in this form. If y(1) = 1 was given, then C = 1, and y = 1/(x − x·ln|x|).
Example 2: Orthogonal Trajectories
Find the orthogonal trajectories of the family of parabolas y = ax².
1
Find the DE of the given family
▶
y = ax². Differentiate: dy/dx = 2ax. But a = y/x², so:
dy/dx = 2(y/x²)·x = 2y/x.
The slope of the tangent at any point on y = ax² is dy/dx = 2y/x.
dy/dx = 2(y/x²)·x = 2y/x.
The slope of the tangent at any point on y = ax² is dy/dx = 2y/x.
2
Orthogonal trajectories have slope = negative reciprocal
▶
For orthogonal trajectories, the slope at any point is the negative reciprocal: dy/dx = −x/(2y).
This is variable separable: 2y·dy = −x·dx
∫ 2y dy = −∫ x dx
y² = −x²/2 + C
This is variable separable: 2y·dy = −x·dx
∫ 2y dy = −∫ x dx
y² = −x²/2 + C
x² + 2y² = 2C → x²/2 + y²/1 = C₁
The orthogonal trajectories are ellipses with semi-axes in ratio √2 : 1. Every parabola y = ax² is intersected orthogonally by these ellipses!
Example 3: Growth/Decay Application
The rate at which a substance decays is proportional to the amount present. If half of a 100g sample decays in 10 years, how much remains after 25 years?
1
Set up and solve the DE
▶
dN/dt = −kN (negative because amount decreases).
Separate: dN/N = −k dt.
Integrate: ln|N| = −kt + C → N = N₀·e^(−kt).
At t = 0, N = 100: N₀ = 100.
So N(t) = 100·e^(−kt).
Separate: dN/N = −k dt.
Integrate: ln|N| = −kt + C → N = N₀·e^(−kt).
At t = 0, N = 100: N₀ = 100.
So N(t) = 100·e^(−kt).
2
Find k from the half-life and compute
▶
Half-life means N = 50 when t = 10:
50 = 100·e^(−10k) → e^(−10k) = ½ → −10k = ln(½) = −ln(2) → k = ln(2)/10 ≈ 0.0693.
At t = 25: N = 100·e^(−25·ln(2)/10) = 100·e^(−2.5·ln(2)) = 100·(e^(ln(2)))⁻²·⁵
= 100·(2⁻²·⁵) = 100/2^(2.5) = 100/(2²·√2) = 100/(4√2) = 25/√2 ≈ 17.68 grams.
50 = 100·e^(−10k) → e^(−10k) = ½ → −10k = ln(½) = −ln(2) → k = ln(2)/10 ≈ 0.0693.
At t = 25: N = 100·e^(−25·ln(2)/10) = 100·e^(−2.5·ln(2)) = 100·(e^(ln(2)))⁻²·⁵
= 100·(2⁻²·⁵) = 100/2^(2.5) = 100/(2²·√2) = 100/(4√2) = 25/√2 ≈ 17.68 grams.
After 25 years: ≈ 17.68 g remaining
🌟 Growth & Decay Formula
dN/dt = ±kN → N(t) = N₀·e^(±kt). | Half-life: t₁/₂ = ln(2)/k
(+) for growth (population, interest), (−) for decay (radioactive, cooling). The magnitude of k determines how fast things change.
Practice Zone: Think, Don't Compute
These questions test if you truly understand differential equations.