Question #a249f

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Question #a249f

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Answer 1 · 2017-06-03T01:40:11

$y = \frac{x^{2}}{4} + 2$ $y=x^2/4+2$

Explanation:

First, integrate both sides with respect to $x$ $x$ .

$\int \frac{d y}{d x} d x = \int \frac{1}{2} x d x$ $intdy/dxdx = int1/2xdx$

$\int d y = \frac{1}{2} \int x d x$ $intdy = 1/2intxdx$

$y + C_{1} = \frac{x^{2}}{4} + C_{2}$ $y + C_1 = x^2/4 + C_2$

$y = \frac{x^{2}}{4} + C_{2} - C_{1}$ $y = x^2/4 + C_2 - C_1$

Let $C = C_{2} - C_{1}$ $C = C_2 - C_1$

$y = \frac{x^{2}}{4} + C$ $y = x^2/4 + C$

Now, to figure out what C is, plug in known $x$ $x$ and $y$ $y$ values and solve for $C$ $C$ .

$3 = \frac{2^{2}}{4} + C$ $3 = 2^2/4 + C$

$3 = 1 + C$ $3 = 1+C$

$2 = C$ $2 = C$

So we can therefore write $y = f (x)$ $y=f(x)$ like this:

$y = \frac{x^{2}}{4} + 2$ $y=x^2/4+2$

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Question #a249f

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