Problem 35 - Advanced Engineering Mathematics

Question

We note that \(\left(M_{y}-N_{x}\right) / N=3,\) so an integrating factor is \(e \int 3 d x=e^{3 x} .\) Let \(M=\left(10-6 y+e^{-3 x}\right) e^{3 x}=\) \(10 e^{3 x}-6 y e^{3 x}+1\) and \(N=-2 e^{3 x},\) so that \(M_{y}=-6 e^{3 x}=N_{x} .\) From \(f_{x}=10 e^{3 x}-6 y e^{3 x}+1\) we obtain \(f=\) \(\frac{10}{3} e^{3 x}-2 y e^{3 x}+x+h(y), h^{\prime}(y)=0,\) and \(h(y)=0 .\) A solution of the differential equation is \(\frac{10}{3} e^{3 x}-2 y e^{3 x}+x=c\)

Step-by-Step Solution

Answer

The solution of the differential equation is \(\frac{10}{3} e^{3x} - 2ye^{3x} + x = c\).

Step 1: Understanding the Integrating Factor

The problem states that an integrating factor is given by \( e^{\int 3 \, dx} = e^{3x} \). This integrating factor is used to make the differential equation exact.

Step 2: Define M and N

The given values are \( M = (10 - 6y + e^{-3x}) e^{3x} = 10e^{3x} - 6ye^{3x} + 1 \) and \( N = -2e^{3x} \). These represent the components of the differential equation our integrating factor will act on.

Step 3: Verify the Condition for Exactness

To verify if the differential equation is exact after applying the integrating factor, calculate \(M_y = -6e^{3x}\) and \(N_x = -6e^{3x}\). Since \(M_y = N_x\), the equation is exact.

Step 4: Integrate M with Respect to x

From \( f_x = 10e^{3x} - 6ye^{3x} + 1 \), integrate with respect to \(x\) to get \( f = \frac{10}{3} e^{3x} - 2ye^{3x} + x + h(y) \). Here, \(h(y)\) is an unknown function of \(y\).

Step 5: Differentiate the Solution with Respect to y

To find \(h(y)\), differentiate \(f\) with respect to \(y\) to ensure it matches \(N\). This gives \(h'(y) = 0\), thus \(h(y) = 0\).

Step 6: Formulate the General Solution

The solution to the differential equation is \(\frac{10}{3} e^{3x} - 2ye^{3x} + x = c\), where \(c\) is a constant obtained after integrating, reflecting the solution form.

Key Concepts

Integrating Factor

When tackling differential equations, one might encounter situations where the equation isn't in a simple, solvable form. The concept of an integrating factor comes in handy in such scenarios. An integrating factor is essentially a function that, when multiplied with a given differential equation, transforms it into an exact equation—a simpler form which is easier to solve. In our exercise, the integrating factor was identified as \( e^{3x} \), derived from \( e^{\int 3 \, dx} \). This factor is calculated by realizing how the terms of the differential equation can be adjusted to meet the criteria of exactness.

Exact Differential Equation

An exact differential equation is a type where certain conditions allow it to be expressible in a straightforward form. For an equation to be exact, the derivatives of its components must be equal in a certain way. In mathematical terms, for a given differential equation \( M(x, y) \ dx + N(x, y) \ dy = 0 \), it is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). By using the integrating factor \( e^{3x} \), our equation in the exercise becomes exact, because it satisfies the criteria after conversion. Hence, verifying this condition was pivotal for ensuring that the method applied was correct.

Integration

In solving exact differential equations, integration plays a crucial role. Once the equation is transformed into an exact form, the next step is to integrate. In our solution, we integrated the component \( M \) with respect to \( x \) to get \( f(x, y) \). Specifically, from \( f_x = 10e^{3x} - 6ye^{3x} + 1 \), integrating with respect to \( x \) resulted in \( \frac{10}{3} e^{3x} - 2ye^{3x} + x + h(y) \). Here, \( h(y) \) stands for any function of \( y \) that might appear from the integration, reflecting that some information is lost during integration, needing further clarification.

Solving Techniques

The overall solving technique for differential equations involves several strategic steps that collectively form a systematic approach. Key steps include finding an integrating factor, checking for exactness, performing integration, and validating the solution. In this exercise, each step was meticulously followed to ensure clarity and accuracy. After applying the integrating factor and ensuring exactness, integration allowed for expressing the general solution. Differentiation ensured that any terms missing due to integration constants were accounted for. Finally, formulating the general solution of \( \frac{10}{3} e^{3x} - 2ye^{3x} + x = c \) represented the culmination of the method, capturing the solution's essence in a constant, \( c \). This example illustrates how these solving techniques create a bridge from a complex differential equation to a comprehensible solution.