Question

Solve the differential equation \( y e^y dx = (y^3 + 2x e^y) dy \), given that \( y(0) = 1 \).

Hint:

When it is difficult to separate variables or formulate a homogeneous equation in terms of \(\frac{dy}{dx}\), invert the derivatives to see if it fits the linear format for \(\frac{dx}{dy}\).

Answer 1

Concept: This differential equation can be rearranged into a linear differential equation of the form: \[ \frac{dx}{dy} + P(y)x = Q(y) \] To solve this, we calculate an Integrating Factor (\(\text{I.F.}\)) given by \( e^{\int P(y) dy} \), and the general solution is found via: \[ x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C \]

Step 1: Rearranging the differential equation to standard linear form.
The given equation is: \[ y e^y dx = (y^3 + 2x e^y) dy \] Rearranging terms to get an expression for \( \frac{dx}{dy} \): \[ \frac{dx}{dy} = \frac{y^3 + 2x e^y}{y e^y} \] Divide each term in the numerator by the denominator: \[ \frac{dx}{dy} = \frac{y^3}{y e^y} + \frac{2x e^y}{y e^y} \quad \Rightarrow \quad \frac{dx}{dy} = y^2 e^{-y} + \frac{2}{y}x \] Bring the term containing \(x\) to the left-hand side: \[ \frac{dx}{dy} - \frac{2}{y}x = y^2 e^{-y} \] This matches the standard form \( \frac{dx}{dy} + P(y)x = Q(y) \), where: \[ P(y) = -\frac{2}{y}, \quad Q(y) = y^2 e^{-y} \]

Step 2: Finding the Integrating Factor (\(\text{I.F.}\)).
The formula for the integrating factor is: \[ \text{I.F.} = e^{\int P(y) dy} = e^{\int -\frac{2}{y} dy} = e^{-2 \ln y} = e^{\ln(y^{-2})} = y^{-2} = \frac{1}{y^2} \]

Step 3: Evaluating the general solution and applying initial conditions.
The general solution template is: \[ x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C \] Substitute \( \text{I.F.} \) and \( Q(y) \): \[ x \cdot \frac{1}{y^2} = \int \left(y^2 e^{-y}\right) \cdot \frac{1}{y^2} dy + C \] \[ \frac{x}{y^2} = \int e^{-y} dy + C \quad \Rightarrow \quad \frac{x}{y^2} = -e^{-y} + C \] We are given that \( y(0) = 1 \), which means \( x = 0 \) when \( y = 1 \): \[ \frac{0}{1^2} = -e^{-1} + C \quad \Rightarrow \quad 0 = -\frac{1}{e} + C \quad \Rightarrow \quad C = \frac{1}{e} = e^{-1} \] Substitute \(C\) back into the general solution: \[ \frac{x}{y^2} = -e^{-y} + e^{-1} \] Multiplying both sides by \(e^y\) directly structures the algebraic format into standard explicit functional variations matching option targets smoothly.