Solution of Ex:2.4 Exact differential equations by Dennis G. Zill

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Solution of Ex:2.4 Exact differential equations by Dennis G. Zill. Exact differential equations are a type of differential equation that can be solved using a specific method known as the method of integrating factors. In the seventh edition of “Differential Equations with Boundary-Value Problems” by Dennis G. Zill, several exact differential equations are solved in exercise 2.4 on page 68. Let’s take a look at some of these solutions.

Differential Equations with BVPs

Overview Examples of Exact DE

The first problem involves finding the solution to the differential equation (x^2 - y) dx - 2xy dy = 0. To check if this equation is exact, we can calculate the partial derivatives of (x^2 - y) with respect to x and y, and the partial derivatives of -2xy with respect to x and y. If these partial derivatives are equal, then the equation is exact. In this case, we have:

    \[\frac{\partial}{\partial y} (x^2 - y) = -1\]


    \[\frac{\partial}{\partial x} (-2xy) = -2y\]

Since these partial derivatives are not equal, the equation is not exact. To make it exact, we * with suitable integrating factor.

Now,

    \[(x^2 - y) e^(-x^2) dx - 2xy e^(-x^2) dy = 0\]

Now, if we check the partial derivatives, we get:

    \[\frac{\partial}{\partial y} [(x^2 - y) e^(-x^2)] = -e^(-x^2)\]


    \[\frac{\partial}{\partial x} [-2xy e^(-x^2)] = -e^(-x^2)\]

Since these partial derivatives are equal, the equation is exact. We can find the potential function by integrating the coefficient of dx with respect to x and taking the derivative with respect to y, or vice versa. In this case, we get:

    \[f(x,y) = ∫(x^2 - y) e^(-x^2) dx + C(y)\]


    \[= -1/2 e^(-x^2) (x^2 + 1) + C(y)\]

Taking the derivative of f with respect to y, we get:

    \[\frac{\partial f}{\partial y} = -C'(y)\]

Comparing this with the coefficient of dy, which is -2xy e^(-x^2), we get:

    \[C'(y) = 2xy e^(-x^2)\]

Integrating both sides with respect to y, we get:

    \[C(y) = x^2 y + D\]

Therefore, the (GS) is:

    \[-1/2 e^(-x^2) (x^2 + 1) + x^2 y + D = 0\]

Another problem involves finding the solution to the differential equation (x^2 + y^2) dx - 2xy dy = 0. In this case, the equation is already exact, since:

    \[\frac{\partial}{\partial y} (x^2 + y^2) = 2y\]


    \[\frac{\partial}{\partial x} (-2xy) = -2y\]

Therefore, we can find the potential function by integrating with respect to x and taking the derivative with respect to y, we get:

    \[f(x,y) = \frac{1}{2}x^2 y + 1/2 y^3\]

Therefore, the general solution (GS) is:

    \[\frac{1}{2} x^2 y + 1/2 y^3 = C\]

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