Question

Solve the following differential equations: (i) \(y x^{y-1} d x+x^{y} \ln x d y=0\) (ii) \(\mathrm{ye}^{-\pi / \mathrm{y}} \mathrm{dx}-\left(\mathrm{xe}^{-\mathrm{x} / \mathrm{y}}+\mathrm{y}^{3}\right) \mathrm{dy}=0\)

Short answer

To summarize, we have solved the given differential equations: 1. For equation (i) \( y x^{y-1} dx + x^y \ln x dy = 0\), the solution is \(y \ln x = -\int \frac{y x^{y-1}}{x\ln x} dx - c\). 2. For equation (ii) \( ye^{-\pi / y}dx - (xe^{-x / y} + y^3) dy = 0\), the solution is \(xy e^{-\pi/y} - \frac{1}{4}y^4 = C\).

Step-by-step solution

Solve the first differential equation #(i) y x^{y-1} dx + x^y \ln x dy = 0.

For this equation, we will use the substitution method. Let \(v = y \ln x\). Then, differentiate \(v\) with respect to \(x\) and use the chain rule: \(v_x = y x^{y-1}.\) Now, we can rewrite the given equation as: \(v_x dx + x^y \ln xd(v) = 0\) Divide the entire equation by \((x\ln x)\): \(\frac{v_x dx}{x\ln x} + \frac{x^y d(v)}{x^y} = 0\) \(\frac{v_x dx}{x\ln x} + d(v) = 0\) Now, integrate both terms: \(\int \frac{v_x}{x\ln x} dx = -\int d(v)\) \(-v = \int \frac{v_x}{x\ln x} dx + c\) Substitute \(v\) back into the equation: \( -y \ln x = \int \frac{y x^{y-1}}{x\ln x} dx + c\) \(y \ln x = -\int \frac{y x^{y-1}}{x\ln x} dx - c\) Now, we have solved equation (i).

Solve the second differential equation #(ii) ye^{-\pi / y}dx - (xe^{-x / y} + y^3) dy = 0.

First, check if the given differential equation is exact by calculating \(M_x\) and \(N_y\): \(M = y e^{-\pi/y}\) and \(N = -x e^{-x/y} - y^3\) \(M_x = -\frac{\pi}{y^2} e^{-\pi/y}\) and \(N_y = \frac{x}{y^2} e^{-x/y} - 3y^2\) Since \(M_x = N_y\), this differential equation is exact. Now, we integrate \(M\) with respect to \(x\), and \(N\) with respect to \(y\): \(\int M dx = \int y e^{-\pi/y} dx = xy e^{-\pi/y} + f(y)\) \(\int N dy = \int (-x e^{-x/y} - y^3) dy = -xy e^{-x/y} - \frac{1}{4}y^4 + g(x)\) Now, to find the potential function \(\phi(x,y)\), we need to find functions \(f(y)\) and \(g(x)\) such that: \(\phi(x,y) = xy e^{-\pi/y} + f(y) = -xy e^{-x/y} - \frac{1}{4}y^4 + g(x)\) The only way \(\phi(x, y)\) can be equal on both sides of the equation is when \(f(y)= - \frac{1}{4}y^4\) and \(g(x) = 0\). Therefore, the potential function is: \(\phi(x,y) = xy e^{-\pi/y} - \frac{1}{4}y^4\) To solve this equation, we set \(\phi(x, y) = C\). Thus, the solution is: \(xy e^{-\pi/y} - \frac{1}{4}y^4 = C\) Now, we have solved equation (ii).