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NSW Y12 Maths - Extension 1 Differential Equations Solving y'=f(x)

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Solving y'=f(x) Theory

Finding a solution to a differential equation means expressing the relationship between the variables in a form which does not contain any derivatives. \\ A simple example is \(\dfrac{d y}{d x}=x^2\), by integrating, we have \(y=\dfrac{x^3}{3}+c\), There is a family of solution curves which all satisfy the differential equation. This is known as the general solution. \\ If given the point \((3,1)\) then \(1=\dfrac{3^3}{3}+C\)\\ \(\therefore c=-8\)\\ \(y=\dfrac{x^3}{3}-8\) is a particular solution.\\ This type of differential equation is of the form \(\dfrac{d y}{d x}=f(x)\).\\ \begin{multicols}{2} \textbf{Example 1}\\%question 118326 Write the general solution to the differential equation \(\dfrac{dy}{dx}=x \sqrt{x}\)\\ \textbf{Example 1 solution}\\ $\begin{aligned} \frac{d y}{d x} &=x \times x^{\frac{1}{2}} \\ \frac{d y}{d x} &=x^{\frac{3}{2}} \\ y &=\int x^{\frac{3}{2}} d x \\ &=\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+c \\ y &=\frac{2}{5} x^{\frac{5}{2}}+c \\ &=\frac{2}{5} \sqrt{x^5}+c \end{aligned}$\\ \columnbreak \textbf{Example 2}\\%question 118328 Find the particular solution to the differential equation \(\dfrac{dy}{dx}=\sqrt{1-x}\) given that when \(x=1, \, y=2\) \\ \textbf{Example 2 solution}\\ $\begin{aligned} \frac{d y}{d x} &=(1-x)^{\frac{1}{2}} \\ y &=\int(1-x)^{\frac{1}{2}} d x \\ &=-\frac{(1-x)^{\frac{3}{2}}}{\frac{3}{2}}+c \\ y &=-\frac{2}{3}(1-x)^{\frac{3}{2}}+c \\ \text{At } (1,2) \quad 2 &=-\frac{2}{3}(1-1)^{\frac{3}{2}}+c \rightarrow c=2 \\ \therefore y &=-\frac{2}{3} \sqrt{(1-x)^3}+2 \end{aligned}$\\ \end{multicols}

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