Find the solution of the initial value problem x2 y' + x y = 1, y(1) = | Mathematics I 2081

Mathematics I208110 marksNumerical: differential equation and areaLinear equations

Find the solution of the initial value problem $x^2 y' + x y = 1$ , $y(1) = 2$ , $x > 0$ .

Find the area enclosed by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$ . [5+5]

Find the solution of the initial value problem $x^2 y' + x y = 1$, $y(1) = 2$, $x > 0$.

Find the area enclosed by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$. [5+5]

--- Given data: $x^2 y' + xy = 1$ , $\; y(1) = 2$ , $\; x 0$ Divide by $x^2$ : $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x^2}$ So $P = \dfrac{1}{x}$ , $Q = \dfrac{1}{x^2}$ . $\text{I.F.} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x$ $$y\cdot x = \int \frac{1}{x^2}\cdot x,dx + C = \int \frac{1}{x},d...

--- Given data: $x^2 y' + xy = 1$, $; y(1) = 2$, $; x 0$ Divide by $x^2$: $$\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x^2}$$ So $P = \dfrac{1}{x}$, $Q = \dfrac{1}{x^2}$. $$\text{I.F.} = e^{\int \frac{1}{x},dx} = e^{\ln x} = x$$ $$y\cdot x = \int \frac{1}{x^2}\cdot x,dx + C = \int \frac{1}{x},d...

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Find the solution of the initial value problem x2 y' + x y = 1, y(1) = 2, x > 0. Find the area enclosed by the line y = x - 1 and the parabola y2 = 2x + 6. [5+5]