Problem 35 - Advanced Engineering Mathematics

Question

Solve the given differential equation by finding, as in Example 4, an appropriate integrating factor. $$ \left(10-6 y+e^{-3 x}\right) d x-2 d y=0 $$

Step-by-Step Solution

Answer

The solution is \(\frac{10}{3}e^{3x} - 6ye^{3x} + x + 4y = C\), where \(C\) is a constant.

Step 1: Identify the equation format

The given equation is \((10 - 6y + e^{-3x})\,dx - 2\,dy = 0\). This needs to be written in the usual differential form \(M\,dx + N\,dy = 0\), where \(M = 10 - 6y + e^{-3x}\) and \(N = -2\).

Step 2: Check if equation is exact

To check if the equation is exact, calculate the partial derivatives: \(\frac{\partial M}{\partial y} = -6\) and \(\frac{\partial N}{\partial x} = 0\). Since \(\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x}\), the equation is not exact.

Step 3: Determine the integrating factor

Using the formula for non-exact equations where \(N(x,y)\, dy\) term has constant partial with respect to \(x\): \(\mu(y) = e^{\int \frac{\partial M/\partial y - \partial N/\partial x}{N} \, dx}\). Here, since \(\frac{\partial M/\partial y - \partial N/\partial x}{N} = \frac{-6}{-2} = 3\), \(\mu(y) = e^{\int 3 \, dx} = e^{3x}\).

Step 4: Multiply by the integrating factor

Multiply the entire differential equation by the integrating factor \(\mu(x) = e^{3x}\): \((e^{3x}(10 - 6y + e^{-3x}))\, dx - 2e^{3x}\, dy = 0\).This simplifies to: \((10e^{3x} - 6ye^{3x} + 1)\, dx - 2e^{3x}\, dy = 0\).

Step 5: Check exactness again

Check if the new equation is exact:\(M = 10e^{3x} - 6ye^{3x} + 1\), \(N = -2e^{3x}\).Calculate \(\frac{\partial M}{\partial y} = -6e^{3x}\) and \(\frac{\partial N}{\partial x} = -6e^{3x}\). Since these are equal, the equation is now exact.

Step 6: Integrate to find the solution

Integrate \(M\) with respect to \(x\) to get: \(F(x, y) = \int (10e^{3x} - 6ye^{3x} + 1)\,dx = \frac{10}{3}e^{3x} - 6ye^{3x} + x + g(y)\) where \(g(y)\) is an arbitrary function of \(y\).

Step 7: Use derivative to find \(g(y)\)

Differentiate \(F\) with respect to \(y\) and equate to \(N\):\(\frac{\partial F}{\partial y} = -6e^{3x} + g'(y) = -2e^{3x}\).Solve for \(g'(y)\): \(g'(y) = 4e^{3x}\).Integrate \(g'(y)\) to get: \(g(y) = 4y\).

Step 8: Write the final solution

Combine the results: \(F(x, y) = \frac{10}{3}e^{3x} - 6ye^{3x} + x + 4y = C\), where \(C\) is a constant. This is the implicit solution to the differential equation.

Key Concepts

Exact Differential Equations

Exact differential equations are a specific type of differential equation where you can identify a function, say \( F(x, y) \), whose total differential \( dF = F_x \, dx + F_y \, dy \) matches the original equation's format \( M \, dx + N \, dy = 0 \).

A differential equation is considered exact if the partial derivatives satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This often implies that there exists a potential function \( F(x, y) \) such that:

• \( M = \frac{\partial F}{\partial x} \)
• \( N = \frac{\partial F}{\partial y} \)

If this holds true, it points to the fact that the original equation is the total differential of some function, making it exact. Knowing this can significantly simplify solving the differential equation, as it then becomes a matter of finding \( F(x, y) \) and setting it equal to a constant.

Non-Exact Differential Equations

Not all differential equations are neatly wrapped into the exact form. Sometimes, when you apply the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), it doesn't hold true. Such equations are non-exact.

For non-exact differential equations, one tool that we often use is the integrating factor. This is a specially chosen function that transforms the non-exact equation into an exact one.

• An integrating factor for the equation \( M \, dx + N \, dy = 0 \) generally depends on either \( x \) or \( y \), or sometimes on both such that multiplication of the whole equation by this function brings about exactness.
• To find the integrating factor, often you will use certain conditions or simplifications, like assuming dependency only on \( x \) or only on \( y \), to ease calculations.

Once you have successfully converted the equation into an exact form using the integrating factor, the solution can proceed as you would with exact equations.

Integration Techniques

Integration is at the heart of solving differential equations, especially when you transform a difficult equation into an exact one.

Once you determine the equation is exact, or have used an integrating factor to make it so, you will often integrate to find the solution. The typical process is:

• Integrate \( M \) with respect to \( x \), treating \( y \) as a constant, which yields part of the potential function \( F(x, y) \).
• This part might include an arbitrary function of \( y \), as it behaves like a constant during the integration with respect to \( x \).
• Next, differentiate the resulting expression with respect to \( y \) to find any additional components or to solve for the arbitrary function based on conditions given by \( N \).

The seamless transitions of integration and differentiation within this method are key to efficiently solving the equations. It ensures that all parts of the potential function are accounted for, leading to the correct implicit or explicit solution of the problem.