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3 years ago

Using the digits 0 to 9 at most one Time each,fill in the boxes below to write equations whose solution is 1/2

Mathematics

1 answer:

andrezito [222]3 years ago

8 0

Answer:

The equation is 4x + 5 = 2x + 6

Step-by-step explanation:

The solution is $\frac{1}{2}$ means x = $\frac{1}{2}$

Let us put the coefficient of x in the left hand side = 4 and the coefficient of x in the right hand side = 2

∵ 4x - 2x = 2x

Now we need two numerical terms their difference is 1 like 5 and 6 or 6 and 7 or 7 and 8 or 8 and 9

Let us take 5 and 6

∵ 4x + 5 = 2x + 6

- Subtract 2x from both sides

∴ 2x + 5 = 6

- Subtract 5 from both sides

∴ 2x = 1

- Divide both sides by 2

∴ x = $\frac{1}{2}$

∴ The solution is $\frac{1}{2}$

The equation is 4x + 5 = 2x + 6

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Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
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So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$