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ch4aika [34]
3 years ago
10

How much must you add to -12 to get a number greater than 5?

Mathematics
1 answer:
Elodia [21]3 years ago
5 0
18 would be 6 so thats the answer
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Read 2 more answers
Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%
BaLLatris [955]
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
8 0
3 years ago
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