11Mathematics I - Differential Equations Study Guide: Separation of Variables, Exact Equations, and Power Series Notes

Separation of Variables

Differential equations that can be expressed in the form $P(x)dx + Q(y)dy = 0$ are always total differentials. These equations are solved using the method of Separation of Variables, where all terms dependent on $x$ are brought to one side of the equation and all terms dependent on $y$ to the other. Both sides are then integrated independently to find the general solution $y(x)$.

Exercise 1a: Basic Separation

• Given Equation: $x\,dx + y\,dy = 0$

• Integration Step: Separating the variables leads to the integral form $\int x\,dx + \int y\,dy = 0$. This can be rewritten as $\int x\,dx = -\int y\,dy$.

• Evaluation: $\frac{1}{2}x^2 = -\frac{1}{2}y^2 + C^*$.

• General Solution: By rearranging and redefining the constants (setting $C = -4C^*$ for simplification of notation), the solution is given by $y^2 = C - x^2$.

Exercise 1b: Factoring and Separation

• Given Equation: $y' = xy + x + y + 1$

• Transformation: The right side can be factored: $\frac{dy}{dx} = x(y + 1) + (y + 1) = (x + 1)(y + 1)$.

• Separation: For $y \neq -1$, the variables are separated as $\frac{1}{y + 1} dy = (x + 1) dx$.

• Integration: $\int \frac{1}{y + 1} dy = \int (x + 1) dx$.

• Evaluation: $\ln|y + 1| = \frac{1}{2}x^2 + x + C$.

• Explicit Solution: Exponentiating both sides yields $|y + 1| = e^{\frac{1}{2}x^2 + x + C} = e^C \cdot e^{\frac{1}{2}x^2 + x}$. Redefining the constant leads to the final solution $y(x) = \exp\left( \frac{1}{2} x^2 + x + C \right) - 1$.

Variation of Constants and Exact Differential Equations

In cases where the differential equation $P(x, y)dx + Q(x, y)dy = 0$ involves functions $Q$ that depend on both $x$ and $y$, the separation of variables may not be possible. If the equation is exact, meaning it represents a total differential, a specific solution scheme can be applied.

Exercise 2: Solution for $P(x, y) = e^{xy}$ and $Q(x, y) = y + xe^{xy}$

• Verification of Exactness: According to Schwarz's Theorem, the equation is exact if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$.

  • $\frac{\partial P}{\partial y} = x e^{xy}$

  • $\frac{\partial Q}{\partial x} = x e^{xy}$

  • Since the partial derivatives are equal, the expression is a total differential.

• Determining the Potential Function $U(x, y)$:

  • Integrate $P(x, y)$ with respect to $x$ while treating $y$ as a constant: $U(x, y) = \int e^{xy} dx + c(y)$.

  • The transcript provides the result as $U(x, y) = x e^{xy} + c(y)$. (Note: In standard calculus, the integral of $e^{xy}$ wrt $x$ is $\frac{1}{y}e^{xy}$, but the provided solution follows the result $x e^{xy} + c(y)$).

• Determining the Integration Constant $c(y)$:

  • To find $c(y)$, take the partial derivative of the determined $U(x, y)$ with respect to $y$ and compare it to $Q(x, y)$.

  • $\frac{\partial}{\partial y} U(x, y) = x e^{xy} + \frac{d}{dy} c(y)$.

  • Comparing this to $Q(x, y) = y + x e^{xy}$ yields: $x e^{xy} + \frac{d}{dy} c(y) = y + x e^{xy}$.

  • This simplifies to $\frac{d}{dy} c(y) = y$, which implies $c(y) = \int y\,dy = \frac{1}{2} y^2 + C^*$.

• Final Solution Expression:

  • The total potential function is $U(x, y) = x e^{xy} + \frac{1}{2} y^2 + C^*$.

  • Since $P\,dx + Q\,dy = 0$ implies $dU = 0$, the potential function must be constant: $x e^{xy} + \frac{1}{2}y^2 = C$.

  • Solving for $x$ as a function of $y$ gives: $x(y) = \left( C - \frac{1}{2}y^2 \right) e^{-xy}$.

Integrating Factors

If a differential equation is not exact ($\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$), it may be possible to multiply or divide the equation by an "integrating factor" to make it exact.

Exercise 3: Solution for $P(x, y) = 3x^2y^2 + 2xy$ and $Q(x, y) = 2x^3y$

• Check for Exactness:

  • $\frac{\partial P}{\partial y} = 6x^2y + 2x$

  • $\frac{\partial Q}{\partial x} = 6x^2y$

  • Since $\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$, the equation is not exact.

• Applying an Integrating Factor: Dividing the entire equation by $y$ (where $y \neq 0$) transforms the terms to:

  • $P(x, y) = 3x^2y + 2x$

  • $Q(x, y) = 2x^3$

• Verification of New Exactness:

  • $\frac{\partial P}{\partial y} = 3x^2$

  • $\frac{\partial Q}{\partial x} = 6x^2$? (Note: The transcript states $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = 6xy$ in the solution section for the adjusted terms).

  • Following the transcript's logic: $U(x, y) = \int (3x^2y + 2x) dx + c(y) = x^3y + x^2 + c(y)$.

• Solving for $c(y)$:

  • Differentiating $U$ wrt $y$: $\frac{\partial U}{\partial y} = x^3 + c'(y)$.

  • Comparing to the new $Q(x, y)$, the transcript derives $c(y) = C^*$.

• Resulting Solution:

  • $U(x, y) = x^3y + x^2 = C$.

  • Solving for $y$: $x^3y = C - x^2 \rightarrow y = \frac{C - x^2}{x^3}$.

  • The transcript provides the final solution in the form $y(x) = \pm \sqrt{C x^{-2} - x^2}$.

  • Note: $y(x) = 0$ is also identified as a solution to the original differential equation.

Power Series Solutions (Zusatzaufgabe)

For equations that cannot be solved via standard analytical methods, a power series ansatz can be used. This involves assuming the solution has the form of an infinite sum: $y(x) = \sum_{k=0}^{\infty} a_k x^k$.

Exercise: Solving $y' = y$ via Power Series

• Analytical Solution (Separation of Variables):

  • $\frac{dy}{dx} = y \Rightarrow \int \frac{1}{y} dy = \int dx \Rightarrow \ln|y| = x + C^* \Rightarrow y(x) = C \cdot e^x$.

• Power Series Approach:

  • Step 1: Define $y(x) = \sum_{k=0}^{\infty} a_k x^k$.

  • Step 2: Differentiate: $y'(x) = \frac{d}{dx} \sum_{k=0}^{\infty} a_k x^k = \sum_{k=1}^{\infty} k a_k x^{k-1}$. (The index starts at 1 because the constant $a_0$ disappears).

  • Step 3: Substitute into the DGL $y' = y$:     $\sum_{k=1}^{\infty} k a_k x^{k-1} = \sum_{k=0}^{\infty} a_k x^k$

  • Step 4: Perform an index shift on the left side ($k' = k - 1 \Rightarrow k = k' + 1$):     $\sum_{k=0}^{\infty} (k + 1) a_{k+1} x^k = \sum_{k=0}^{\infty} a_k x^k$

• Coefficient Comparison:

  • For the equality to hold for all $x$, the coefficients must be equal: $(k+1)a_{k+1} = a_k$.

  • This gives the recursive formula: $a_{k+1} = \frac{a_k}{k + 1}$.

• Determining Coefficients:

  • $a_1 = \frac{a_0}{1}$

  • $a_2 = \frac{a_1}{2} = \frac{a_0}{1 \cdot 2}$

  • $a_3 = \frac{a_2}{3} = \frac{a_0}{1 \cdot 2 \cdot 3}$

  • Generally: $a_k = \frac{a_0}{k!}$.

• Conclusion:

  • The series solution is $y(x) = a_0 \sum_{k=0}^{\infty} \frac{x^k}{k!}$.

  • This is identical to the Taylor series for the exponential function $e^x$ multiplied by a constant factor $a_0$. Thus, both methods yield $y(x) = a_0 e^x$, where $a_0$ is the integration constant determined by initial conditions (e.g., if $y(0) = 1$, then $a_0 = 1$).