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NSW Y12 Maths - Extension 1 Differential Equations Introduction to Differential Equations (DE's)

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Introduction to Differential Equations (DE's) Theory

\textbf{Classifying differential equations (DE's)} \\ \(\dfrac{d y}{d x}=k x\) is a first order, first degree \(D E\).\\ This is because it involves a first order derivative to a power of one. \\ \(\left(\dfrac{d y}{d x}\right)^2=k x\) is a first order, second degree. \(D E\). This is because it involves a first order derivative to a power of two.\\ \(\dfrac{d^2 y}{d x^2}+k \dfrac{d y}{d x}+c=0\) is a second order, first degree \(D E\). This is because it involves a second order derivative to a power of one.\\ \begin{multicols}{2} \textbf{Example 1}\\ %question 123245 Classify the differential equation \(x^2 \dfrac{d y}{d x}+1=0\)\\ \textbf{Example 1 solution}\\ \(x^2 \dfrac{d y}{d x}+1=0\) defines a first -order, first degree differential equation because it involves a first order derivative to a power of one.\\ \columnbreak \textbf{Example 2}\\%question 123248 Given that \(y=5 x^3\) is a solution to \(x y^{\prime}-k y=0 \text {, then } k=\text { ? }\)\\ \textbf{Example 2 solution}\\ $\begin{aligned} y=&5 x^3 \\ y^{\prime}=&15 x^2 \end{aligned}$\\ $\begin{aligned} x \times 15 x^2-k \times 5 x^3&=0 \\ 15 x^3-5 k x^3&=0 \\ (15-5 k) x^3&=0 \\ \therefore 15-5 k&=0 \\ \therefore k&=3 \end{aligned}$ \end{multicols}

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  • Introduction to Differential Equations (DEs) - Video - Solving differential equations

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