Complete Differential Equations Guide

Interactive solutions with step-by-step explanations and visualizations

1. Automobile Velocity Problem

M(dV/dt) = F - kV, with V(0) = 0

1 Separate variables:

Rearrange the equation to separate V and t:

dV/(F - kV) = dt/M

2 Integrate both sides:

∫dV/(F - kV) = ∫dt/M

This gives:

(-1/k)ln|F - kV| = t/M + C

3 Apply initial condition:

At t=0, V=0:

(-1/k)ln|F| = C

4 Solve for V:

Combine and exponentiate:

F - kV = Fe^(-kt/M)

Then:

V = (F/k)(1 - e^(-kt/M))

Velocity vs Time

Final Solution

V(t) = (F/k)(1 - e^(-kt/M))

This shows the velocity approaches F/k asymptotically as t increases.

2. Parachute Velocity Problem

v(dv/dx) = g(1 - v²/k²), with v(0) = 0

1 Separate variables:

Rearrange the equation:

v/(1 - v²/k²) dv = g dx

2 Integrate both sides:

∫v/(1 - v²/k²) dv = ∫g dx

Use substitution u = 1 - v²/k², du = -2v/k² dv:

(-k²/2)∫du/u = gx + C

3 Solve the integral:

(-k²/2)ln|1 - v²/k²| = gx + C

4 Apply initial condition:

At x=0, v=0:

C = (-k²/2)ln|1| = 0

5 Solve for v:

Exponentiate both sides:

1 - v²/k² = e^(-2gx/k²)

Then:

v = k√(1 - e^(-2gx/k²))

Velocity vs Distance Fallen

Final Solution

v(x) = k√(1 - e^(-2gx/k²))

This shows the velocity approaches terminal velocity k as x increases.

3. Curve Equation Problem

dy/dx = (y - 1)/(x² + x), passes through (1,0)

1 Separate variables:

Rearrange the equation:

dy/(y - 1) = dx/(x² + x)

2 Factor denominator:

x² + x = x(x + 1)

Use partial fractions:

1/(x(x+1)) = 1/x - 1/(x+1)

3 Integrate both sides:

∫dy/(y - 1) = ∫(1/x - 1/(x+1))dx

This gives:

ln|y - 1| = ln|x| - ln|x + 1| + C

4 Combine logarithms:

ln|y - 1| = ln|x/(x + 1)| + C

5 Exponentiate both sides:

y - 1 = K(x/(x + 1)) (where K = eᴰ)

6 Apply initial condition:

At x=1, y=0:

-1 = K(1/2) ⇒ K = -2

Thus:

y = 1 - 2x/(x + 1)

Curve Visualization

Final Solution

y(x) = 1 - 2x/(x + 1) = (1 - x)/(1 + x)

This is the equation of the curve passing through (1,0) with the given slope.

4(i). Separable Equation

dy/dx = √[(1 - y²)/(1 - x²)]

1 Separate variables:

Rearrange the equation:

dy/√(1 - y²) = dx/√(1 - x²)

2 Integrate both sides:

∫dy/√(1 - y²) = ∫dx/√(1 - x²)

These are standard integrals:

sin⁻¹(y) = sin⁻¹(x) + C

3 General solution:

We can write:

y = sin(sin⁻¹(x) + C)

Or using trigonometric identity:

y = x cos C + √(1 - x²) sin C

Solution Curves

Final Solution

y = sin(sin⁻¹(x) + C) or y = x cos C + √(1 - x²) sin C

This represents a family of solutions parameterized by C.

4(ii). Exact Equation

y dx + (1 + x²) tan⁻¹x dy = 0

1 Separate variables:

Rearrange the equation:

dy/y = -dx/[(1 + x²) tan⁻¹x]

2 Integrate both sides:

∫dy/y = -∫dx/[(1 + x²) tan⁻¹x]

For the right side, let u = tan⁻¹x, du = dx/(1 + x²):

ln|y| = -∫du/u = -ln|u| + C

3 Solve for y:

Exponentiate both sides:

y = eᴰ/(tan⁻¹x) = K/(tan⁻¹x)

Solution Curves

Final Solution

y = K/(tan⁻¹x)

Where K is an arbitrary constant.

4(iii). Unusual Differential Equation

sin(dy/dx) = a, with y(0) = 1

1 Solve for dy/dx:

dy/dx = sin⁻¹(a) + 2πn or π - sin⁻¹(a) + 2πn

Where n is any integer.

2 Integrate:

The simplest solution (n=0):

y = x sin⁻¹(a) + C

or

y = x(π - sin⁻¹(a)) + C

3 Apply initial condition:

At x=0, y=1:

C = 1

Solution Curves

Final Solution

y = x sin⁻¹(a) + 1

or

y = x(π - sin⁻¹(a)) + 1

There are infinitely many solutions corresponding to different branches of arcsin.

4(iv). Exponential Differential Equation

dy/dx = eˣ⁺ʸ + x³eʸ = eʸ(eˣ + x³)

1 Separate variables:

Rearrange the equation:

e⁻ʸ dy = (eˣ + x³) dx

2 Integrate both sides:

∫e⁻ʸ dy = ∫(eˣ + x³) dx

This gives:

-e⁻ʸ = eˣ + x⁴/4 + C

3 Solve for y:

e⁻ʸ = -eˣ - x⁴/4 - C

Take natural log of both sides:

y = -ln(-eˣ - x⁴/4 - C)

Solution Curves

Final Solution

y = -ln(-eˣ - x⁴/4 - C)

Where C must be chosen such that the argument of ln is positive.

(v). Exact Differential Equation

(eʸ + 1)cosx dx + eʸ sinx dy = 0

1 Rearrange equation:

(eʸ + 1)cosx dx = -eʸ sinx dy

2 Separate variables:

Divide both sides by (eʸ + 1)sinx:

cosx/sinx dx = -eʸ/(eʸ + 1) dy

3 Simplify:

cotx dx = -eʸ/(eʸ + 1) dy

4 Integrate both sides:

∫cotx dx = -∫eʸ/(eʸ + 1) dy

This gives:

ln|sinx| = -ln|eʸ + 1| + C

5 Combine logarithms:

ln|sinx(eʸ + 1)| = C

6 Final solution:

Exponentiate both sides:

sinx(eʸ + 1) = K (where K = eᴰ)

Solution Curves

Final Solution

sinx(eʸ + 1) = K

Where K is an arbitrary constant.

(vi). Homogeneous Differential Equation

(ydx - xdy)cot(x/y) = ny²dx

1 Divide by y²:

(dx/y - xdy/y²)cot(x/y) = ndx

2 Recognize differential:

Note that d(x/y) = (ydx - xdy)/y²:

d(x/y)cot(x/y) = ndx

3 Rewrite cotangent:

cos(x/y)/sin(x/y) d(x/y) = ndx

4 Integrate both sides:

Let u = x/y:

∫cosu/sinu du = n∫dx

This gives:

ln|sinu| = nx + C

5 Final solution:

Substitute back u = x/y:

sin(x/y) = Keⁿˣ

Solution Curves

Final Solution

sin(x/y) = Keⁿˣ

Where K is an arbitrary constant.

(vii). Simple Separable Equation

dy/dx = x√(25 - x²)

1 Separate variables:

dy = x√(25 - x²) dx

2 Integrate both sides:

∫dy = ∫x√(25 - x²) dx

Use substitution u = 25 - x², du = -2x dx:

y = -½∫√u du = -⅓u^(3/2) + C

3 Substitute back:

y = -⅓(25 - x²)^(3/2) + C

Solution Curves

Final Solution

y = -⅓(25 - x²)^(3/2) + C

Where C is an arbitrary constant.

(viii). Exact Differential Equation

x cosy dy = eˣ(x logx + 1) dx

1 Separate variables:

Divide both sides by x:

cosy dy = eˣ(logx + 1/x) dx

2 Integrate both sides:

∫cosy dy = ∫eˣ(logx + 1/x) dx

Notice that d/dx(eˣlogx) = eˣlogx + eˣ/x:

siny = eˣlogx + C

Solution Curves

Final Solution

siny = eˣlogx + C

Where C is an arbitrary constant.

(ix). Trigonometric Differential Equation

tany dy/dx = cos(x + y) + cos(x - y)

1 Simplify right side:

Using trigonometric identity:

cos(x+y) + cos(x-y) = 2cosx cosy

So the equation becomes:

tany dy/dx = 2cosx cosy

2 Rewrite tangent:

(siny/cosy) dy/dx = 2cosx cosy

3 Separate variables:

Multiply both sides by cosy/siny:

dy/siny = 2cosx dx

4 Integrate both sides:

∫csc y dy = 2∫cosx dx

This gives:

ln|cscy - coty| = 2sinx + C

Solution Curves

Final Solution

ln|cscy - coty| = 2sinx + C

Where C is an arbitrary constant.

(x). Substitution Differential Equation

dy/dx = tan²(x + y)

1 Substitution:

Let u = x + y, then du/dx = 1 + dy/dx:

du/dx - 1 = tan²u

2 Rearrange:

du/dx = 1 + tan²u = sec²u

3 Separate variables:

cos²u du = dx

4 Integrate both sides:

∫cos²u du = ∫dx

Using cos²u = (1 + cos2u)/2:

½u + ¼sin2u = x + C

5 Substitute back:

Replace u with x + y:

½(x + y) + ¼sin(2x + 2y) = x + C

Simplify:

½y + ¼sin(2x + 2y) = ½x + C

Solution Curves

Final Solution

½y + ¼sin(2x + 2y) = ½x + C

Where C is an arbitrary constant.