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2 years ago

Differential cos^2x dy/dx =e^y-tanx

Mathematics

1 answer:

gulaghasi [49]2 years ago

8 0

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

+y=tanx

⇒

+ysec

x=tanxsec

x ....(1)

Here P=sec

x⇒∫PdP=∫sec

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

tanx

ysec

x=e

tanx

tanxsec

Integrating both sides, we get

tanx

=∫e

tanx

.tanxsec

xdx

Put tanx=t⇒sec

xdx=dt

∴ye

=∫te

dt=e

(t−1)+c

⇒y=t−1+ce

−t

where t=tanx

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Answer:

Step-by-step explanation:

To sketch a quadratic function we need two things:

1) Nature of the curve

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3) y-intercept

The completing square form of the quadratic equation is:

$y=a(x-h)^2+k$

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h represents the x-coordinate of the vertex.

k represents the y-coordinate of the vertex.

Now if we compare g(x) with our completing square form we get the following:

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When we simply compare the following we get ,

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k = 12, which means the y-coordinate of the vertex is 12

Now we have the nature of the curve, we have the vertex now all we need is the y-intercept.

For y-intercept:

For y-intercept meaning at which point will the graph cross the y-axis(0 , y)

For that we expand the formula and turn it into the standard quadratic equation form by using the formula (a - b)^2

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now we compare with the standard quadratic form:

$y=ax^2+bx+c$

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so the curve cuts the y-axis at (0 , 28)

So we have all the three things that we need to graph our function.

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