Given that y=2 when x=-3,What is the value of y when x=5 for the differential equation y dy/dx =2x+3?

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Answer 1 · 2018-06-28T15:03:52

$y^{2} = 2 x^{2} + 6 x + 4$ $y^2 = 2x^2 +6x + 4$

$x = 5 \Rightarrow y = \pm 2 \sqrt{21}$ $x=5 => y = +- 2sqrt(21)$

We have:

$y \frac{d y}{d x} = 2 x + 3$ $ydy/dx=2x+3$ with $y (- 3) = 2$ $y(-3)=2$

The DE is separable, so we simply separate the variables" to get:

$int \ y \ dy = int \ 2x+3 \ dx$

And Integrating we get:

$1/2y^2 = x^2 +3x + C$

Using the initial condition, $y(-3)=2$ , we have:

$1/2(2^2) = (-3)^2 +3(-3) + C$
$\ \ \ \ \ => 2 = 9 - 9 + C$
$\ \ \ \ \ => C=2$

Thus, we gain the particular solution:

$1/2y^2 = x^2 +3x + 2$
$\ \ \ \ \ => y^2 = 2x^2 +6x + 4$

And when $x=5$ we have:

$y^2 = 2(5^2) +6(5) + 4$
$\ \ \ \ =50 +30 + 4$
$\ \ \ \ = 84$

And so there are two solutions:

$y=+-sqrt(84) = +- 2sqrt(21)$