Question #431e3

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Question #431e3

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Answer 1 · 2017-05-26T04:35:53

$e^{2 x} y + \frac{3}{2} = \frac{3 e^{2 x}}{2}$ $e^(2x)y+3/2=(3e^(2x))/2$

Explanation:

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

Can be expressed as:
$\frac{d y}{d x} + y f (x) = g (x)$ $dy/dx+yf(x)=g(x)$

Where $f (x) = 2$ $f(x)=2$ and $g (x) = 3$ $g(x)=3$ .

Finding the Integrating Factor:
$\int f (x) d x = \int 2 d x = 2 x$ $int f(x) dx=int 2 dx=2x$

Raise $e$ $e$ by this: $e^{2 x}$ $e^(2x)$
Multiply the original expression by this factor:
$\frac{d y}{d x} (e^{2 x}) + 2 e^{2 x} y = 3 e^{2 x}$ $dy/(dx)(e^(2x))+2e^(2x)y=3e^(2x)$

Observing the left hand side it is just the Product Rule applied.
Which means it can be rewritten as this:
$(e^(2x)y)'=3e^(2x)$

Integrate both sides
$int(e^(2x)y)'=int3e^(2x)$
$e^(2x)y+C=(3e^(2x))/2$

Find $C$ given the condition $f(0)=0$ .
$e^(2(0))(0)+C=3/2(e^(2(0)))$
$C=3/2$

Putting it all together:
$e^(2x)y+3/2=(3e^(2x))/2$

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Question #431e3

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