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ankoles [38]
3 years ago
9

Consider the implicit differential equation

20%2898%20xy%5E%7B2%7D%20%2B50%20x%5E%7B2%7D%20%29dy%3D0" id="TexFormula1" title="(49 y^{3} + 45 xy) dx + (98 xy^{2} +50 x^{2} )dy=0" alt="(49 y^{3} + 45 xy) dx + (98 xy^{2} +50 x^{2} )dy=0" align="absmiddle" class="latex-formula">. For the integrating factor x^{p}  y^{q} of this equation, find p and q. Now multiply the equation by the integrating factor x^{p}  y^{q} that you have found and then integrate the resulting equation to get a solution in implicit form.
Mathematics
1 answer:
BaLLatris [955]3 years ago
8 0
We're looking for an integrating factor \mu(x,y)=x^py^q such that

\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0

is exact, which would require that

(\mu M)_y=(\mu N)_x
(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x
49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q
\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4

You can verify that (\mu M)_y=(\mu N)_x if you'd like. With the ODE now exact, we have a solution F(x,y)=C such that

F_x=\mu M
F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx
F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)

F_y=\mu N
50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4
f'(y)=0
\implies f(y)=C

and so the general solution is

F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C
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7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7BW%7D%20%3D%202" id="TexFormula1" title="\sqrt[4]{W} = 2" alt="\sqrt[4]{W} = 2
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Answer:

<h2>W = 16</h2>

Step-by-step explanation:

<h3>\sqrt[4]{W}  = 2 </h3>

To find W raise each of the sides of the equation to the power 4 to make W stand alone

That's

<h3>( { \sqrt[4]{W} })^{4}  =  {2}^{4}</h3>

We have

W = 2⁴

We have the final answer as

<h3>W = 16</h3>

Hope this helps you

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