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3 years ago

10

Round 4.9987 to the nearest thousand

1 answer:

lina2011 [118]3 years ago

3 0

Since it is to the nearest thousand, it would be 4.999

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In the figure, p || q.

notsponge [240]

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3 years ago

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Hitman42 [59]

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2 years ago

Select the statement in if-then form.

SashulF [63]

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3 years ago

Solve the differential equation dy dx + 12x2y = 36x2. SOLUTION The given equation is linear since it has the form of this equati

Georgia [21]

Answer:

Therefore the solution of the differential equation is

$ye^{4x^3} = 3e^{4x^3}+c$ [ where c is arbitrary constant]

Step-by-step explanation:

Given differential equation is

$\frac{dy}{dx} +12x^2y= 36x^2$

Here $P(x)= 12x^2$ and $Q(x) = 36 x^2$

The integrating factor of the differential equation is

$= e^{\int P(x) dx$

$=e^{\int 12x^2dx$

$=e^{ \frac{12x^3}{3}}$

$=e^{4x^3}$

Multiplying the integrating factor both sides of the differential equation

$e^{4x^3}\frac{dy}{dx} +12x^2ye^{4x^3}= 36x^2e^{4x^3}$

$\Rightarrow e^{4x^3} dy+12x^2ye^{4x^3}dx= 36x^2e^{4x^3}dx$

Integrating both sides,

$\int e^{4x^3} dy+\int12x^2ye^{4x^3}dx= \int36x^2e^{4x^3}dx$ ......(1)

Let

$I= \int36x^2e^{4x^3}dx$

$= \int3. 12 x^2e^{4x^3}dx$

putting ${4x^3}=z$ , $12x^2 dx=dz$

$=\int 3. e^zdz$

$=3e^z+c$ [ where c is arbitrary constant]

Putting the value of z

$=3e^{4x^3}+c$

From (1) we get

$\int e^{4x^3} dy+\int12x^2ye^{4x^3}dx= \int36x^2e^{4x^3}dx$

$\Rightarrow ye^{4x^3} = 3e^{4x^3}+c$

Therefore the solution of the differential equation is

$ye^{4x^3} = 3e^{4x^3}+c$ [ where c is arbitrary constant]

7 0

4 years ago