Question

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

Short answer

Question: For the exact differential equation, find the value of \(b\) and solve the equation for that value: $$ \left(ye^{2xy} + x\right) dx + bx e^{2xy} dy = 0 $$ Answer: The value of \(b\) for which the given equation is exact is \(b = 1\). The solution to the exact differential equation is: $$ \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} = C $$

Step-by-step solution

Determine the value of b for which the given equation is exact

To find the value of \(b\), we need to check for the exactness of the given equation. For a differential equation of the form \(M(x, y) dx + N(x, y) dy = 0\), the necessary condition for exactness is: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Let's find the partial derivatives for our given equation: $$ M(x,y) = ye^{2xy} + x \\ N(x,y) = bx e^{2xy} $$ Therefore, $$ \frac{\partial M}{\partial y} = \frac{\partial (ye^{2xy} + x)}{\partial y} = e^{2xy} + 2xye^{2xy} \\ \frac{\partial N}{\partial x} = \frac{\partial (bx e^{2xy})}{\partial x} = b e^{2xy} + 2bxy e^{2xy} $$ Now, equate \(\frac{\partial M}{\partial y}\) with \(\frac{\partial N}{\partial x}\): $$ e^{2xy} + 2xye^{2xy} = b e^{2xy} + 2bxy e^{2xy} $$ Since \(e^{2xy}\) doesn't equal zero, we can divide both sides of the equation by it: $$ 1 + 2xy = b + 2bxy $$ We can see that \(1 = b\) and \(2xy = 2xy\). Therefore, the value of \(b\) for which the given equation is exact is \(b = 1\).

Solve the exact differential equation using the value of b

Now that we have found the value of \(b\), we can rewrite our given exact differential equation as: $$ \left(ye^{2xy} + x\right) dx + xe^{2xy} dy = 0 $$ As the equation is exact, there exists a function \(\psi(x, y)\) such that \(\frac{\partial \psi}{\partial x} = M(x, y)\) and \(\frac{\partial \psi}{\partial y} = N(x, y)\). Let's find \(\psi(x, y)\) using this condition. Integrate \(\frac{\partial \psi}{\partial x}\) with respect to \(x\): $$ \psi(x, y) = \int (ye^{2xy} + x) dx $$ This integration gives: $$ \psi(x, y) = \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} + g(y) $$ Now we need to find the function \(g(y)\). To do that, differentiate \(\psi(x, y)\) with respect to \(y\): $$ \frac{\partial \psi}{\partial y} = \frac{\partial (\frac{1}{2} x^2 + \frac{1}{2}e^{2xy} + g(y))}{\partial y} = xe^{2xy} + g'(y) $$ We know that \(\frac{\partial \psi}{\partial y} = N(x, y) = xe^{2xy}\). Comparing the expressions, we get \(g'(y)=0\) which implies that \(g(y)\) is a constant, let's say \(C\). Our final solution will be in the form \(\psi(x, y) = C\): $$ \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} + C = 0 $$ So, the solution to the given exact differential equation for \(b = 1\) is: $$ \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} = C $$