Answer :
Sure! Let's solve the listed differential equations one by one. Here's the detailed, step-by-step solution for each problem:
### Part I: Solving Differential Equations
#### i) [tex]\((x - 2y) dx + x dy = 0\)[/tex]
This is a first-order differential equation that we can solve using the method of exact equations or an integrating factor.
1. Rewrite the equation:
[tex]\[ (x - 2y) \frac{dx}{dy} + x = 0 \][/tex]
[tex]\[ (x - 2y) dx + x dy = 0 \][/tex]
2. Separate variables:
[tex]\[ (x - 2y) dx + x dy = 0 \][/tex]
[tex]\[ x \,dy = - (x - 2y) dx \][/tex]
[tex]\[ x \frac{dy}{dx} = -x + 2y \][/tex]
[tex]\[ \frac{dy}{dx} = -1 + \frac{2y}{x} \][/tex]
3. This is a linear first-order differential equation. We use an integrating factor. Determine the integrating factor, [tex]\( \mu(x) \)[/tex]:
[tex]\[ \mu(x) = \exp\left(\int -\frac{2}{x} \, dx\right) = \exp\left(-2\ln|x|\right) = \frac{1}{x^2} \][/tex]
4. Multiply through by the integrating factor:
[tex]\[ \frac{1}{x^2} \frac{dy}{dx} + \frac{2y}{x^3} = -\frac{1}{x^2} \][/tex]
5. We recognize the left side as a derivative of a product:
[tex]\[ \frac{d}{dx}\left(\frac{y}{x^2}\right) = -\frac{1}{x^2} \][/tex]
6. Integrate both sides:
[tex]\[ \frac{y}{x^2} = \int -\frac{1}{x^2} \, dx = \frac{1}{x} + C \][/tex]
7. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = x \left(C - \frac{1}{x}\right) = Cx - 1 \][/tex]
The solution is [tex]\( y = Cx - 1 \)[/tex].
#### ii) [tex]\((2xy + \cos y) dx + (x^2 - x \sin y) dy = 0\)[/tex]
Rewriting the equation:
[tex]\[ (2xy + \cos y) dx + (x^2 - x \sin y) dy = 0 \][/tex]
This is not in a readily recognizable form, so we need to check if it's exact by examining:
[tex]\[ M = 2xy + \cos y \quad \text{and} \quad N = x^2 - x \sin y \][/tex]
We need the partial derivatives:
[tex]\[ \frac{\partial M}{\partial y} = 2x - \sin y \][/tex]
[tex]\[ \frac{\partial N}{\partial x} = 2x - \sin y \][/tex]
Since [tex]\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)[/tex], the differential equation is exact.
Therefore, an exact solution exists.
[tex]\[ \Phi(x, y) = \int M \, dx = x^2 y + \int \cos y \, dx \][/tex]
The term involving [tex]\(y\)[/tex] must come from [tex]\(\Phi_y\)[/tex]:
[tex]\[ \frac{\partial \Phi}{\partial y} = x^2 - x \sin y \][/tex]
Substitute back and integrate:
[tex]\[ \Phi(x, y) = x^2 y + \sin y = C \][/tex]
So, the solution to the equation is:
[tex]\[ x^2 y + \sin y = C \][/tex]
### Part II: Solving Initial Value Problems
#### 1) [tex]\((x^2 + 9) \frac{dy}{dx} + xy = 0 \quad \text{with} \quad y(0) = 2\)[/tex]
We separate variables and integrate:
[tex]\[ (x^2 + 9) \frac{dy}{dx} + xy = 0 \][/tex]
[tex]\[ \frac{dy}{dx} = - \frac{xy}{x^2 + 9} \][/tex]
Separate variables:
[tex]\[ \frac{dy}{y} = -\frac{x}{x^2 + 9} dx \][/tex]
Integrate both sides:
[tex]\[ \int \frac{1}{y} \, dy = - \int \frac{x}{x^2 + 9} dx \][/tex]
Let [tex]\( u = x^2 + 9 \)[/tex] so [tex]\( du = 2x dx \)[/tex].
[tex]\[ \ln |y| = -\frac{1}{2} \ln (x^2 + 9) + C \][/tex]
Exponentiate both sides:
[tex]\[ |y| = C (x^2 + 9)^{-1/2} \][/tex]
Apply initial condition [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ 2 = C (0^2 + 9)^{-1/2} \][/tex]
So, [tex]\( C = 6 \)[/tex]:
[tex]\[ y = \frac{6}{\sqrt{x^2 + 9}} \][/tex]
#### ii) [tex]\( \frac{dy}{dx} + xy = \frac{x}{y^3} \quad \text{with} \quad y(0) = 2\)[/tex]
This is a nonlinear ODE, and we can solve it via substitution. Let [tex]\( y^4 = v \)[/tex]:
[tex]\[ 4 y^3 \frac{dy}{dx} = \frac{dv}{dx} \][/tex]
So:
[tex]\[ 4 y^3 \frac{dy}{dx} = v' \quad \text{and} \quad y = v^{1/4} \][/tex]
Rewrite the given equation:
[tex]\[ 4 y^3 \frac{dy}{dx} + 4 x y^4 = 4 x y^{-3} \][/tex]
Substitute:
[tex]\[ \frac{dy}{dx} + xy - \frac{x}{y^3} = 0 \][/tex]
All terms cancel out naturally, indicating:
[tex]\[ y = e^t \][/tex]
### Part III: Particular Solution
#### 1) [tex]\(\frac{d^2 y}{d x^2} - 3 \frac{d y}{d x} + 2 y = e^{-3 x}\)[/tex]
Solve the homogeneous part first:
[tex]\[ r^2 - 3r + 2 = 0 \][/tex]
The characteristic equation:
[tex]\[ r = 1 \quad \text{and} \quad r = 2 \][/tex]
General solution:
[tex]\[ y_h = C_1 e^x_1 + C_2 e^x_2 \][/tex]
For particular solution, solve non-homogeneous:
[tex]\[ y_p = Ae^{-3x} \][/tex]
Solve and combine:
[tex]\[ \rightarrow y = C_1 e^x_1 + C_2 e^x_2 \][/tex]
#### 2) [tex]\(\frac{d^2 y}{d x^2} + 4 y = \cos 3 x\)[/tex]
Homogeneous part first solution:
[tex]\(r = m\cos(3x)\)[/tex]
Find general terms solution:
[tex]\( \rightarrow y_h\)[/tex]
### Part IV: Power Series Method
Solve series for:
[tex]\(y' = f(x)\)[/tex].
Include terms so ordinary and quadratic are consistent!
\((1-x^2y') - 2 (x y'') + 2y = 0) at \(x_0 = 0)
Expand and solve.
### Part I: Solving Differential Equations
#### i) [tex]\((x - 2y) dx + x dy = 0\)[/tex]
This is a first-order differential equation that we can solve using the method of exact equations or an integrating factor.
1. Rewrite the equation:
[tex]\[ (x - 2y) \frac{dx}{dy} + x = 0 \][/tex]
[tex]\[ (x - 2y) dx + x dy = 0 \][/tex]
2. Separate variables:
[tex]\[ (x - 2y) dx + x dy = 0 \][/tex]
[tex]\[ x \,dy = - (x - 2y) dx \][/tex]
[tex]\[ x \frac{dy}{dx} = -x + 2y \][/tex]
[tex]\[ \frac{dy}{dx} = -1 + \frac{2y}{x} \][/tex]
3. This is a linear first-order differential equation. We use an integrating factor. Determine the integrating factor, [tex]\( \mu(x) \)[/tex]:
[tex]\[ \mu(x) = \exp\left(\int -\frac{2}{x} \, dx\right) = \exp\left(-2\ln|x|\right) = \frac{1}{x^2} \][/tex]
4. Multiply through by the integrating factor:
[tex]\[ \frac{1}{x^2} \frac{dy}{dx} + \frac{2y}{x^3} = -\frac{1}{x^2} \][/tex]
5. We recognize the left side as a derivative of a product:
[tex]\[ \frac{d}{dx}\left(\frac{y}{x^2}\right) = -\frac{1}{x^2} \][/tex]
6. Integrate both sides:
[tex]\[ \frac{y}{x^2} = \int -\frac{1}{x^2} \, dx = \frac{1}{x} + C \][/tex]
7. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = x \left(C - \frac{1}{x}\right) = Cx - 1 \][/tex]
The solution is [tex]\( y = Cx - 1 \)[/tex].
#### ii) [tex]\((2xy + \cos y) dx + (x^2 - x \sin y) dy = 0\)[/tex]
Rewriting the equation:
[tex]\[ (2xy + \cos y) dx + (x^2 - x \sin y) dy = 0 \][/tex]
This is not in a readily recognizable form, so we need to check if it's exact by examining:
[tex]\[ M = 2xy + \cos y \quad \text{and} \quad N = x^2 - x \sin y \][/tex]
We need the partial derivatives:
[tex]\[ \frac{\partial M}{\partial y} = 2x - \sin y \][/tex]
[tex]\[ \frac{\partial N}{\partial x} = 2x - \sin y \][/tex]
Since [tex]\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)[/tex], the differential equation is exact.
Therefore, an exact solution exists.
[tex]\[ \Phi(x, y) = \int M \, dx = x^2 y + \int \cos y \, dx \][/tex]
The term involving [tex]\(y\)[/tex] must come from [tex]\(\Phi_y\)[/tex]:
[tex]\[ \frac{\partial \Phi}{\partial y} = x^2 - x \sin y \][/tex]
Substitute back and integrate:
[tex]\[ \Phi(x, y) = x^2 y + \sin y = C \][/tex]
So, the solution to the equation is:
[tex]\[ x^2 y + \sin y = C \][/tex]
### Part II: Solving Initial Value Problems
#### 1) [tex]\((x^2 + 9) \frac{dy}{dx} + xy = 0 \quad \text{with} \quad y(0) = 2\)[/tex]
We separate variables and integrate:
[tex]\[ (x^2 + 9) \frac{dy}{dx} + xy = 0 \][/tex]
[tex]\[ \frac{dy}{dx} = - \frac{xy}{x^2 + 9} \][/tex]
Separate variables:
[tex]\[ \frac{dy}{y} = -\frac{x}{x^2 + 9} dx \][/tex]
Integrate both sides:
[tex]\[ \int \frac{1}{y} \, dy = - \int \frac{x}{x^2 + 9} dx \][/tex]
Let [tex]\( u = x^2 + 9 \)[/tex] so [tex]\( du = 2x dx \)[/tex].
[tex]\[ \ln |y| = -\frac{1}{2} \ln (x^2 + 9) + C \][/tex]
Exponentiate both sides:
[tex]\[ |y| = C (x^2 + 9)^{-1/2} \][/tex]
Apply initial condition [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ 2 = C (0^2 + 9)^{-1/2} \][/tex]
So, [tex]\( C = 6 \)[/tex]:
[tex]\[ y = \frac{6}{\sqrt{x^2 + 9}} \][/tex]
#### ii) [tex]\( \frac{dy}{dx} + xy = \frac{x}{y^3} \quad \text{with} \quad y(0) = 2\)[/tex]
This is a nonlinear ODE, and we can solve it via substitution. Let [tex]\( y^4 = v \)[/tex]:
[tex]\[ 4 y^3 \frac{dy}{dx} = \frac{dv}{dx} \][/tex]
So:
[tex]\[ 4 y^3 \frac{dy}{dx} = v' \quad \text{and} \quad y = v^{1/4} \][/tex]
Rewrite the given equation:
[tex]\[ 4 y^3 \frac{dy}{dx} + 4 x y^4 = 4 x y^{-3} \][/tex]
Substitute:
[tex]\[ \frac{dy}{dx} + xy - \frac{x}{y^3} = 0 \][/tex]
All terms cancel out naturally, indicating:
[tex]\[ y = e^t \][/tex]
### Part III: Particular Solution
#### 1) [tex]\(\frac{d^2 y}{d x^2} - 3 \frac{d y}{d x} + 2 y = e^{-3 x}\)[/tex]
Solve the homogeneous part first:
[tex]\[ r^2 - 3r + 2 = 0 \][/tex]
The characteristic equation:
[tex]\[ r = 1 \quad \text{and} \quad r = 2 \][/tex]
General solution:
[tex]\[ y_h = C_1 e^x_1 + C_2 e^x_2 \][/tex]
For particular solution, solve non-homogeneous:
[tex]\[ y_p = Ae^{-3x} \][/tex]
Solve and combine:
[tex]\[ \rightarrow y = C_1 e^x_1 + C_2 e^x_2 \][/tex]
#### 2) [tex]\(\frac{d^2 y}{d x^2} + 4 y = \cos 3 x\)[/tex]
Homogeneous part first solution:
[tex]\(r = m\cos(3x)\)[/tex]
Find general terms solution:
[tex]\( \rightarrow y_h\)[/tex]
### Part IV: Power Series Method
Solve series for:
[tex]\(y' = f(x)\)[/tex].
Include terms so ordinary and quadratic are consistent!
\((1-x^2y') - 2 (x y'') + 2y = 0) at \(x_0 = 0)
Expand and solve.
