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Questions and Answers
What is the general form of a homogeneous differential equation?
Which method is commonly used to solve first-order linear ordinary differential equations?
In an initial value problem, what is typically given?
When using the separation of variables method, what is the first step?
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What is the outcome of solving the equation $2yy' - (y')^2 = 0$?
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Which condition defines a differential equation as homogeneous?
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What substitution is typically used for simplifying homogeneous differential equations when $M(x, y)$ is simpler?
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In the process of separating variables, how is the equation manipulated?
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When integrating with respect to $x$ after separating variables, which technique is useful?
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In the homogenous equation $ye^{2x}dx = (4 + e^{2x})dy$, which method is applied to separate the variables?
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Upon solving the initial value problem using $@ x = 2, y = 3$, what is the value of the constant $C$ determined in the solution?
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What is the form of the final solution for the equation $y^2 + 1 = Cx$ after solving the initial value problem?
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What is the primary purpose of using integrating factors in differential equations?
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What is the general form of a second-order linear homogeneous differential equation?
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Which of the following methods is NOT typically used for solving homogeneous differential equations?
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In the substitution $y = e^{2x}$ for the equation $y''' - 3y' + 2y = 0$, which derived value is incorrectly calculated?
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What must be satisfied for the solution $y = C_1 ext{cos}( heta t) + C_2 ext{sin}( heta t)$ to be valid for the equation $rac{d^2x}{dt^2} + heta^2 x = 0$?
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Which of the following represents the characteristic equation derived from $y'' - 3y' + 2y = 0$?
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When applying the method of elimination of arbitrary constants, what is typically required?
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During the verification process of $e^{2x}$ as a solution to a differential equation, what does substituting yield?
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What expression is incorrect for the term $y'''$ for the function $y = C_1e^{3x} + C_2e^{-2x}$?
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Study Notes
Variable Separable Differential Equation
- A differential equation is said to be variable separable if all terms containing x and y can be grouped with dx and dy respectively.
Types of Solutions
- General solution: Contains an arbitrary constant that is not yet defined.
- Particular solution: Contains no arbitrary constant. The arbitrary constant has been defined.
How to Solve Variable Separable Differential Equations
- Group terms with x and dx on one side, and terms with y and dy on the other side.
- Integrate both sides after grouping.
- The solution will usually contain an arbitrary constant, which can be determined if initial conditions are provided.
Examples
-
Example 1: $y=2x$
- The general solution is $y=xe^c$, where c is an arbitrary constant.
-
Example 2: $2x(y+1)dx-ydy=0$
- The particular solution, given the initial condition x = 1, y = 0, is $x^2 - y -ln(y+1) + 1 = c$.
Homogeneous Differential Equations
- A differential equation is homogeneous if every term has the same degree.
- It can be written in the form $M(x, y)dx + N(x, y)dy = 0$, where M and N are functions of x and y with the same degree.
- Homogeneous differential equations can be solved by making a suitable substitution.
How to Solve Homogeneous Differential Equations
- If M is simpler, substitute $x = vy$ and $dx = vdy + ydx$.
- If N is simpler, substitute $y = vx$ and $dy = vdx + xdv$.
Elimination of Arbitrary Constant
- The elimination of arbitrary constant allows us to obtain the differential equation from a given equation.
- We differentiate the given equation based on the number of arbitrary constants.
Elimination of Arbitrary Constant Using Elimination Method (Determinant)
- This method involves constructing a determinant with the given equation and its derivatives.
- The determinant is then set to zero to obtain the differential equation.
Examples
-
Example 1: $y = C_1 + C_2e^{3x}$
- The differential equation is $y'' - 3y' = 0$.
-
Example 2: $2xyy' = 1 + y^2$
- The particular solution, given the initial condition x = 2, y = 3, is $y^2 + 1 = 5x$.
-
Example 3: $(x^3 + 2x)y dx - x(y^4 + 3)y dy = 0$
- The particular solution, given the initial condition x = 1, y = 1, is $\frac{1}{3}x^3 + 2ln(x) = \frac{1}{4}y^4 + 3ln(y) + \frac{1}{12}$.
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Example 4: $ye^{2x}dx = (4+e^{2x})dy$
- The general solution is $4 + e^{2x} = Cy^2$, where C is an arbitrary constant.
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Description
Explore the concept of variable separable differential equations with this quiz. Learn how to differentiate between general and particular solutions, and solve examples using integration. Test your understanding with practical applications of these principles.