DIFFERENTIAL EQUATIONS (With Solutions)

DIFFERENTIAL EQUATIONS

1. Solve the following Differential Equations

(a) \( \frac{dy}{dx} + 2y = xy^2 \)

(b) \( 3\frac{dy}{dx} + \frac{3y}{x} = 2x^4y^4 \)

2. Solve the following Differential Equations

(a) \( x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = 0 \)

(b) \( \frac{d^2y}{dx^2} = 2\left(\frac{dy}{dx}\right)^2 \)

3. Find the general solution of the following Differential Equations

(a) \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \)

(b) \( y'' + 2y' + 2y = \cos x \)

(c) \( y'' - 8y' + 16y = 8\sin 2x + 3e^{4x} \)

4. Population Growth Problem

Suppose that the population of a colony of bacteria increases exponentially. At the start of an experiment, there are 600 bacteria, and one hour later, the population has increased to 640. How long will it take for the population to reach 1000?

5. Time of Death Estimation

For a postmortem, a doctor requires to know approximately the time of death of the diseased. He recorded the first temperature at 10:00 a.m to be 93.4°F. After 2 hours, he finds the temperature to be 91.4°F. If the room temperature (which is constant) is 72°F, estimate the time of death, assuming normal temperature of human body to be 98.04°F.

6. Additional Problems

(a) Solve \( x^2\frac{dy}{dx} = y^2 - xy + y^2 \) given that \( y(1) = 1 \)

(b) (i) Formulate a differential equation from \( y = A\sinh nx + B\cosh nx \)
(ii) Solve the equation \( (2x^3 + 3y)dx + (3x + y - 1)dy = 0 \)

(c) Find the general solution of \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = e^{2x} \)

(d) The acceleration of the body is directly proportional to the displacement \( x \). When the displacement is 20m, the acceleration of the body is \( 5m/s^2 \). Determine the equation of the path when \( t = 0, \frac{dx}{dt} = 1m/s \) and \( x = 8m \).

DIFFERENTIAL EQUATIONS SOLUTIONS

1. Solve the following Differential Equations

(a) \( \frac{dy}{dx} + 2y = xy^2 \)

This is a Bernoulli equation of the form \( \frac{dy}{dx} + P(x)y = Q(x)y^n \)

1. Divide both sides by \( y^2 \): \( y^{-2}\frac{dy}{dx} + 2y^{-1} = x \)

2. Let \( v = y^{-1} \), then \( \frac{dv}{dx} = -y^{-2}\frac{dy}{dx} \)

3. Substitute: \( -\frac{dv}{dx} + 2v = x \) ⇒ \( \frac{dv}{dx} - 2v = -x \)

4. Integrating factor: \( \mu(x) = e^{\int -2 dx} = e^{-2x} \)

5. Multiply through: \( e^{-2x}\frac{dv}{dx} - 2e^{-2x}v = -xe^{-2x} \)

6. Integrate both sides: \( v e^{-2x} = \int -xe^{-2x} dx \)

Using integration by parts, we get: \( v e^{-2x} = \frac{x}{2}e^{-2x} + \frac{1}{4}e^{-2x} + C \)

7. Solve for v: \( v = \frac{x}{2} + \frac{1}{4} + Ce^{2x} \)

8. Recall \( v = y^{-1} \): \( y = \frac{1}{\frac{x}{2} + \frac{1}{4} + Ce^{2x}} \)

Final Answer: \( y = \frac{4}{2x + 1 + 4Ce^{2x}} \)

(b) \( 3\frac{dy}{dx} + \frac{3y}{x} = 2x^4y^4 \)

This is another Bernoulli equation

1. Divide by \( 3y^4 \): \( y^{-4}\frac{dy}{dx} + \frac{y^{-3}}{x} = \frac{2}{3}x^4 \)

2. Let \( v = y^{-3} \), \( \frac{dv}{dx} = -3y^{-4}\frac{dy}{dx} \)

3. Substitute: \( -\frac{1}{3}\frac{dv}{dx} + \frac{v}{x} = \frac{2}{3}x^4 \)

4. Multiply by -3: \( \frac{dv}{dx} - \frac{3}{x}v = -2x^4 \)

5. Integrating factor: \( \mu(x) = e^{\int -\frac{3}{x} dx} = x^{-3} \)

6. Multiply through: \( x^{-3}\frac{dv}{dx} - 3x^{-4}v = -2x \)

7. Integrate: \( vx^{-3} = \int -2x dx = -x^2 + C \)

8. Solve for v: \( v = -x^5 + Cx^3 \)

9. Recall \( v = y^{-3} \): \( y^{-3} = -x^5 + Cx^3 \)

Final Answer: \( y = \frac{1}{(-x^5 + Cx^3)^{1/3}} \)

2. Solve the following Differential Equations

(a) \( x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} = 0 \)

This is a Cauchy-Euler equation

1. Assume solution of form \( y = x^r \)

2. Compute derivatives: \( \frac{dy}{dx} = rx^{r-1} \), \( \frac{d^2y}{dx^2} = r(r-1)x^{r-2} \)

3. Substitute into equation: \( x^2[r(r-1)x^{r-2}] - 2x[rx^{r-1}] = 0 \)

4. Simplify: \( r(r-1)x^r - 2rx^r = 0 \) ⇒ \( [r^2 - r - 2r]x^r = 0 \)

5. Characteristic equation: \( r^2 - 3r = 0 \) ⇒ \( r(r-3) = 0 \)

6. Roots: \( r = 0 \), \( r = 3 \)

Final Answer: \( y = C_1 + C_2x^3 \)

(b) \( \frac{d^2y}{dx^2} = 2\left(\frac{dy}{dx}\right)^2 \)

1. Let \( v = \frac{dy}{dx} \), then \( \frac{d^2y}{dx^2} = \frac{dv}{dx} \)

2. Substitute: \( \frac{dv}{dx} = 2v^2 \)

3. Separate variables: \( \frac{dv}{v^2} = 2 dx \)

4. Integrate: \( -\frac{1}{v} = 2x + C_1 \)

5. Solve for v: \( v = \frac{-1}{2x + C_1} \)

6. Recall \( v = \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{-1}{2x + C_1} \)

7. Integrate again: \( y = -\frac{1}{2}\ln|2x + C_1| + C_2 \)

Final Answer: \( y = -\frac{1}{2}\ln|2x + C_1| + C_2 \)

3. Find the general solution of the following Differential Equations

(a) \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \)

This is a second-order linear homogeneous ODE with constant coefficients

1. Characteristic equation: \( ar^2 + br + c = 0 \)

2. Roots: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Case 1: Distinct real roots (\( b^2 > 4ac \)): \( y = C_1e^{r_1x} + C_2e^{r_2x} \)

Case 2: Repeated real roots (\( b^2 = 4ac \)): \( y = (C_1 + C_2x)e^{rx} \) where \( r = -\frac{b}{2a} \)

Case 3: Complex roots (\( b^2 < 4ac \)): Let \( r = \alpha \pm i\beta \), then \( y = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x) \)

Final Answer: Solution depends on discriminant \( b^2 - 4ac \) as shown above

(b) \( y'' + 2y' + 2y = \cos x \)

This is a nonhomogeneous linear ODE

1. Solve homogeneous equation: \( y'' + 2y' + 2y = 0 \)

Characteristic equation: \( r^2 + 2r + 2 = 0 \) ⇒ \( r = -1 \pm i \)

Homogeneous solution: \( y_h = e^{-x}(C_1\cos x + C_2\sin x) \)

2. Find particular solution \( y_p \) using undetermined coefficients

Assume \( y_p = A\cos x + B\sin x \)

Compute derivatives and substitute into original equation

After solving, we find \( A = 0 \), \( B = \frac{1}{3} \)

Thus \( y_p = \frac{1}{3}\sin x \)

Final Answer: \( y = e^{-x}(C_1\cos x + C_2\sin x) + \frac{1}{3}\sin x \)

(c) \( y'' - 8y' + 16y = 8\sin 2x + 3e^{4x} \)

1. Solve homogeneous equation: \( y'' - 8y' + 16y = 0 \)

Characteristic equation: \( r^2 - 8r + 16 = 0 \) ⇒ \( r = 4 \) (double root)

Homogeneous solution: \( y_h = (C_1 + C_2x)e^{4x} \)

2. Find particular solutions for each term:

For \( 8\sin 2x \): Assume \( y_{p1} = A\cos 2x + B\sin 2x \)

For \( 3e^{4x} \): Since \( e^{4x} \) is in homogeneous solution, assume \( y_{p2} = Cx^2e^{4x} \)

After solving, we get: \( y_{p1} = \frac{8}{25}\cos 2x + \frac{16}{25}\sin 2x \), \( y_{p2} = \frac{3}{2}x^2e^{4x} \)

Final Answer: \( y = (C_1 + C_2x)e^{4x} + \frac{8}{25}\cos 2x + \frac{16}{25}\sin 2x + \frac{3}{2}x^2e^{4x} \)

4. Population Growth Problem

Given exponential growth model: \( \frac{dP}{dt} = kP \)

1. General solution: \( P(t) = P_0e^{kt} \)

2. At t=0: \( P(0) = 600 = P_0 \)

3. At t=1: \( P(1) = 640 = 600e^{k} \)

4. Solve for k: \( e^{k} = \frac{640}{600} = \frac{16}{15} \) ⇒ \( k = \ln\left(\frac{16}{15}\right) \)

5. Find t when P(t) = 1000: \( 1000 = 600e^{kt} \)

6. Solve: \( e^{kt} = \frac{5}{3} \) ⇒ \( t = \frac{\ln(5/3)}{k} = \frac{\ln(5/3)}{\ln(16/15)} \)

Final Answer: Approximately 11.77 hours

5. Time of Death Estimation

Newton's Law of Cooling: \( \frac{dT}{dt} = k(T - T_{\text{room}}) \)

1. General solution: \( T(t) = T_{\text{room}} + (T_0 - T_{\text{room}})e^{kt} \)

2. Given: \( T_{\text{room}} = 72°F \), first measurement at t=0: \( T(0) = 93.4 \)

At t=2: \( T(2) = 91.4 \)

3. Set up equations: \( 93.4 = 72 + (T_0 - 72)e^{k·0} \) ⇒ \( T_0 = 93.4 \) \( 91.4 = 72 + (93.4 - 72)e^{2k} \)

4. Solve for k: \( e^{2k} = \frac{19.4}{21.4} \) ⇒ \( k = \frac{1}{2}\ln\left(\frac{19.4}{21.4}\right) \)

5. Find time when T(t) = 98.04 (normal body temp): \( 98.04 = 72 + (93.4 - 72)e^{kt} \)

6. Solve for t: \( t = \frac{\ln\left(\frac{26.04}{21.4}\right)}{k} \)

Final Answer: Approximately 2.2 hours before first measurement, so death occurred around 7:48 a.m.

6. Additional Problems

(a) Solve \( x^2\frac{dy}{dx} = y^2 - xy + x^2 \) given that \( y(1) = 1 \)

This is a homogeneous equation

1. Rewrite: \( \frac{dy}{dx} = \left(\frac{y}{x}\right)^2 - \frac{y}{x} + 1 \)

2. Let \( v = \frac{y}{x} \), \( y = vx \), \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

3. Substitute: \( v + x\frac{dv}{dx} = v^2 - v + 1 \)

4. Separate variables: \( \frac{dv}{v^2 - 2v + 1} = \frac{dx}{x} \)

5. Integrate: \( \int \frac{dv}{(v-1)^2} = \int \frac{dx}{x} \) ⇒ \( -\frac{1}{v-1} = \ln|x| + C \)

6. Solve for v: \( v = 1 - \frac{1}{\ln|x| + C} \)

7. Recall \( y = vx \): \( y = x\left(1 - \frac{1}{\ln|x| + C}\right) \)

8. Apply initial condition y(1)=1: \( 1 = 1\left(1 - \frac{1}{0 + C}\right) \) ⇒ \( C = 1 \)

Final Answer: \( y = x\left(1 - \frac{1}{\ln|x| + 1}\right) \)

(b)

(i) Formulate a differential equation from \( y = A\sinh nx + B\cosh nx \)

1. Compute first derivative: \( y' = An\cosh nx + Bn\sinh nx \)

2. Compute second derivative: \( y'' = An^2\sinh nx + Bn^2\cosh nx = n^2y \)

Final Answer: \( y'' - n^2y = 0 \)

(ii) Solve \( (2x^3 + 3y)dx + (3x + y - 1)dy = 0 \)

Check if exact: \( \frac{\partial M}{\partial y} = 3 \), \( \frac{\partial N}{\partial x} = 3 \)

Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact

1. Find F(x,y) such that \( \frac{\partial F}{\partial x} = 2x^3 + 3y \) and \( \frac{\partial F}{\partial y} = 3x + y - 1 \)

2. Integrate first equation: \( F = \frac{1}{2}x^4 + 3xy + g(y) \)

3. Differentiate and compare: \( 3x + g'(y) = 3x + y - 1 \) ⇒ \( g'(y) = y - 1 \)

4. Integrate: \( g(y) = \frac{1}{2}y^2 - y + C \)

Final Answer: \( \frac{1}{2}x^4 + 3xy + \frac{1}{2}y^2 - y = C \)

(c) Find the general solution of \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = e^{2x} \)

1. Solve homogeneous equation: \( y'' - 2y' + y = 0 \)

Characteristic equation: \( r^2 - 2r + 1 = 0 \) ⇒ \( r = 1 \) (double root)

Homogeneous solution: \( y_h = (C_1 + C_2x)e^x \)

2. Find particular solution: Assume \( y_p = Ae^{2x} \)

Substitute and solve: \( A = 1 \)

Final Answer: \( y = (C_1 + C_2x)e^x + e^{2x} \)

(d) The acceleration is proportional to displacement: \( \frac{d^2x}{dt^2} = kx \)

1. Given when x=20m, a=5m/s²: \( 5 = k·20 \) ⇒ \( k = 0.25 \)

2. Solve \( \frac{d^2x}{dt^2} - 0.25x = 0 \)

Characteristic equation: \( r^2 - 0.25 = 0 \) ⇒ \( r = \pm 0.5 \)

General solution: \( x(t) = C_1e^{0.5t} + C_2e^{-0.5t} \)

3. Apply initial conditions: At t=0: \( x(0) = 8 = C_1 + C_2 \) \( \frac{dx}{dt} = 0.5C_1e^{0.5t} - 0.5C_2e^{-0.5t} \) At t=0: \( \frac{dx}{dt} = 1 = 0.5C_1 - 0.5C_2 \)

4. Solve system: \( C_1 = 5 \), \( C_2 = 3 \)

Final Answer: \( x(t) = 5e^{0.5t} + 3e^{-0.5t} \)