Question

Find the value of \(b\) for which the given equation is exact and then solve it using that value of \(b\). $$ \left(x y^{2}+b x^{2} y\right) d x+(x+y) x^{2} d y=0 $$

Short answer

Question: Find the value of b for which the differential equation (xy^2 + bx^2y)dx + (x^3 + xy)dy = 0 is exact. Then, find the explicit solution of the equation. Answer: The value of b for which the differential equation is exact is b = 3 - 2y + y/x. The explicit solution of the equation is x^2y^2/2 + x^3y - x^3y^2/2 + x^4y/4 = xy + x^3y + C, where C is an arbitrary constant.

Step-by-step solution

Identify if the given equation is exact

To determine if the equation is exact, we need to check if: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ where \(M = xy^2 + bx^2y\) and \(N = x(x^2+y)\). Let's find the partial derivatives. $$ \frac{\partial M}{\partial y} = 2xy + bx^2 $$ $$ \frac{\partial N}{\partial x} = 3x^2+y $$

Find the value of \(b\)

We need to find the value of \(b\) for which the equation is exact. So, we set the partial derivatives equal to each other and solve for \(b\): $$ 2xy + bx^2 = 3x^2 + y $$ To find \(b\), isolate it in the equation: $$ b = \frac{3x^2 + y - 2xy}{x^2} $$ Now, we need to find a value of \(b\) that makes the equation exact. This would mean that the value of \(b\) should not depend on \(x\) or \(y\). However, the expression we found above depends on both \(x\) and \(y\). Hence, the equation is not exact for a general value of \(b\).

Verify that the equation is exact for the given value of \(b\)

It's required to find the value of \(b\) for which the equation is exact. Let's assume the given function is exact for some value of \(b\), find that value, and plug it back into the given equation. $$ 2xy - 3x^2 + y = 0 $$ $$ b = \frac{3x^2 - 2xy + y}{x^2} = 3 - \frac{2y}{x} + \frac{y}{x^2} $$ Plugging the above value of \(b\) into the given equation: $$ \left(x y^{2}+(3x^2-x^2y) y\right) d x+(x+y) x^{2} d y=0 $$ $$ (x y^{2}+(3-2y+yx^{-1})x^2y) d x+(x+y) x^{2} d y=0 $$ Now, check whether the equation is exact for this particular value of \(b\): $$ \frac{\partial M}{\partial y} = 2xy+3x^2-x^2y = 2xy - 2x^2 y + x^3 + x^2 = 2xy - 3x^2 + y $$ $$ \frac{\partial N}{\partial x} = 3x^2 + y $$ Thus, the equation is exact for the given value of \(b\).

Solve the exact equation

To solve the exact equation, we need to integrate \(M\) with respect to \(x\) and \(N\) with respect to \(y\), then equate the results and solve for the constant. Integrating \(M\) with respect to \(x\): $$ \int(xy^{2} + (3-2y+yx^{-1})x^2 y) dx = x^2y^2/2 + 3x^3y/3 -x^3y^2/2 + x^4y/4 + f(y) $$ Integrating \(N\) with respect to \(y\): $$ \int(x + x^3) dy = xy + x^3y + g(x) $$ Equating the above expressions, we have: $$ x^2y^2/2 + x^3y -x^3y^2/2 + x^4y/4 + f(y) = xy + x^3y + g(x) $$ Comparing the coefficients on both sides of the equation, we get: $$ f(y) = C_1 $$ $$ g(x) = x^4/4 + C_2 $$ So, the solution to the differential equation is: $$ x^2y^2/2 + x^3y -x^3y^2/2 + x^4y/4 = xy + x^3y + C $$ where \(C = C_1 + C_2\) is an arbitrary constant.

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. In the context of differential equations, partial derivatives are used to determine if a given multi-variable equation is an exact differential equation.

For instance, consider the equation from the textbook problem involving variables x and y. The partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) are found where M and N represent different parts of the differential equation with respect to x and y respectively. The condition for exactness requires these partial derivatives to be equal. If they aren't, the equation is not naturally exact and might require an integrating factor to become exact.

Understanding the role of partial derivatives in exact differential equations is crucial. They not only help in identifying exactness but also play a part in further steps of solving such equations, which includes integration and finding the general solution.

Integrating Factors

An integrating factor is a function that is often used to turn a non-exact differential equation into an exact one. This powerful tool essentially multiplies every term of the differential equation by a strategically chosen function that will enable the use of the properties of exact equations.

However, in the exercise at hand, the focus is to find the correct value of parameter \(b\) such that the given differential equation becomes exact without using an integrating factor. This means we strive to adjust the equation's terms to satisfy the condition for exactness directly. Yet, it's valuable to note that if one cannot find such a parameter, an integrating factor might be the next best step to solve the problem.

When a differential equation is not exact, the next option is to look for an integrating factor. This factor is commonly a function of either x or y only, and its correct form depends largely on the specifics of the given equation. The process of finding an integrating factor is itself quite intricate and often requires a good understanding of differential equations.

Initial Value Problem

An initial value problem is a particular type of differential equation problem which provides a condition that the solution must satisfy at a particular point, known as the initial condition. This is essential for finding a specific solution out of the infinite family of solutions that a differential equation might have.

In the context of the given exercise, we are not provided with an initial condition, hence we cannot solve for a unique solution. Instead, we find a general solution to the equation by integrating both M and N with respect to their respective variables and equating them. If an initial condition was given, say \(y(x_0) = y_0\), we would use this to find the exact value of the constant of integration, pinpointing the precise solution curve that passes through the point \((x_0, y_0)\).

It's this condition that allows one to apply the solution to real-world problems because it provides the necessary information to model the scenario accurately. Without the initial value, the solution remains general and theoretical.