Differential Equations

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These are the solutions to Mathematics questions on Differential Equations.
The TI-84 Plus CE shall be used for applicable questions.
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Formulas and Notes

$ (1.) \text{ Linear First-Order Differential Equation for x in terms of y} \\[3ex] \dfrac{dx}{dy} + P(y)x = Q(y) \\[5ex] \mu(y) = e^{\displaystyle\int P(y)dy} \\[5ex] where \\[3ex] P(y)x \text{ is a function of } x, y \\[3ex] Q(y) \text{ is a function of } y \\[3ex] \mu(y) \text{ is the integrating factor} \\[3ex] $

(1.) Solve the differential equation $\dfrac{dy}{dx} = \dfrac{y}{x + y^3}$ for y > 0

$ \dfrac{dy}{dx} = \dfrac{y}{x + y^3} \\[5ex] \text{Solving in this form "as is" is challenging} \\[3ex] \text{Express as a Linear First-Order Differential Equation} \\[3ex] \dfrac{dx}{dy} = \dfrac{x + y^3}{y} \\[5ex] \dfrac{dx}{dy} = \dfrac{x}{y} + \dfrac{y^3}{y} \\[5ex] \dfrac{dx}{dy} - \dfrac{x}{y} = y^2...eqn.(1) \\[5ex] \text{Compare to: } \dfrac{dx}{dy} + P(y)x = Q(y) \\[5ex] where:\;\; P(y)x = -\dfrac{x}{y} \hspace{2em}and\hspace{2em} Q(y) = y^2 \\[5ex] \implies \\[3ex] P(y) = -\dfrac{1}{y} \\[5ex] \text{integrating factor} = \mu(y) = e^{\displaystyle\int P(y)dy} \\[5ex] \mu(y) = e^{\displaystyle\int -\dfrac{1}{y}dy} \\[5ex] = e^{-\ln |y|}...\text{Because } y \gt 0, \text{ discard the absolute value} \\[4ex] = e^{\ln(y^{-1})} \\[4ex] = y^{-1} \\[3ex] = \dfrac{1}{y} \\[5ex] \mu(y) * eqn.(1) \implies \\[3ex] \dfrac{1}{y}\left(\dfrac{dx}{dy} - \dfrac{x}{y}\right) = \dfrac{1}{y} * y^2 \\[5ex] \left(\dfrac{1}{y}\right)\left(\dfrac{dx}{dy}\right) - \dfrac{x}{y^2} = y \\[5ex] ................................................................................. \\[3ex] \left(\dfrac{1}{y}\right)\left(\dfrac{dx}{dy}\right) \\[5ex] \text{Let us find a relationship between the LHS and the integrating factor} \\[3ex] Let\;\; y = x * \mu(y) \\[3ex] y = x * \dfrac{1}{y} \\[5ex] y = \dfrac{x}{y} \\[5ex] y = x * \dfrac{1}{y} \\[5ex] \dfrac{dy}{dx} = x * -\dfrac{1}{y^2} + \dfrac{1}{y} * \dfrac{dx}{dy}...\text{Product Rule} \\[5ex] \dfrac{dy}{dx} = -\dfrac{x}{y^2} + \left(\dfrac{1}{y}\right)\left(\dfrac{dx}{dy}\right) \\[5ex] \implies \\[3ex] \left(\dfrac{1}{y}\right)\left(\dfrac{dx}{dy}\right) - \dfrac{x}{y^2} \\[5ex] = LHS \\[3ex] = \dfrac{dy}{dx}\;\;of\;\;\mu(y) \\[5ex] = \dfrac{dy}{dx}\;\;of\;\;\dfrac{x}{y} \\[5ex] = \dfrac{d}{dx}\left(\dfrac{x}{y}\right) \\[5ex] ................................................................................. \\[3ex] \text{Substitute for the LHS} \\[3ex] \dfrac{x}{y} = y \\[5ex] \text{Integrate both sides wrt y} \\[3ex] \displaystyle\int \dfrac{d}{dx}\left(\dfrac{x}{y}\right) dy = \displaystyle\int y dy \\[5ex] \dfrac{x}{y} = \dfrac{y^2}{2} + C \\[5ex] x = y\left(\dfrac{y^2}{2} + C\right) \\[5ex] x = \dfrac{y^3}{2} + Cy $

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