Solve: y' = fracx2y2, y(0) = 2. Solve the initial value problem: y'' + | Mathematics I 2078

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Mathematics I207810 marksNumerical: solving differential equationsIntroduction to first order equations Separable equations

Solve: $y' = \frac{x^2}{y^2}$ , $y(0) = 2$ .

Solve the initial value problem: $y'' + y' - 6y = 0$ , $y(0) = 0$ , $y'(0) = 1$ . [5+5]

Solve: $y' = \frac{x^2}{y^2}$, $y(0) = 2$.

Solve the initial value problem: $y'' + y' - 6y = 0$, $y(0) = 0$, $y'(0) = 1$. [5+5]

Part 1: ODE $y' = \dfrac{x^2}{y^2}$ , initial condition $y(0)=2$ . Part 2: ODE $y'' + y' - 6y = 0$ , initial conditions $y(0)=0$ , $y'(0)=1$ . --- $\frac{dy}{dx} = \frac{x^2}{y^2} \implies y^2 \, dy = x^2 \, dx$ $\int y^2 \, dy = \int x^2 \, dx$ $\frac{y^3}{3} = \frac{x^3}{3} + C1$ Multiply by 3...

Part 1: ODE $y' = \dfrac{x^2}{y^2}$, initial condition $y(0)=2$. Part 2: ODE $y'' + y' - 6y = 0$, initial conditions $y(0)=0$, $y'(0)=1$. --- $$\frac{dy}{dx} = \frac{x^2}{y^2} \implies y^2 , dy = x^2 , dx$$ $$\int y^2 , dy = \int x^2 , dx$$ $$\frac{y^3}{3} = \frac{x^3}{3} + C1$$ Multiply by 3...

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Solve: y' = fracx2y2, y(0) = 2. Solve the initial value problem: y'' + y' - 6y = 0, y(0) = 0, y'(0) = 1. [5+5]