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Lady_Fox [76]
3 years ago
15

Which of the following is the correct expansion of (3x-y)2

Mathematics
1 answer:
Studentka2010 [4]3 years ago
3 0

\bold{[ \ Answer \ ]}

\boxed{\bold{9x^{2} \ - \ 6xy \ + y^{2} }}

\bold{[ \ Explanation \ ]}

  • \bold{Find \ Expansion: \ (3x \ - \ y^{2} )}

\bold{----------------------}

  • \bold{[Apply \ Perfect \ Square \ Formula\ (\left(a-b\right)^2=a^2-2ab+b^2)] \ a \ = \ 3x, \ b \ = \ y}

\bold{\left(3x\right)^2-2\cdot \:3xy+y^2}

  • \bold{[Simplify \ \left(3x\right)^2-2\cdot \:3xy+y^2] \ 9x^2-6xy+y^2}

= \bold{9x^2-6xy+y^2}

\boxed{\bold{[] \ Eclipsed \ []}}

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(X^2+y^2+x)dx+xydy=0<br> Solve for general solution
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Check if the equation is exact, which happens for ODEs of the form

M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0

if \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.

We have

M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y

N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y

so the ODE is not quite exact, but we can find an integrating factor \mu(x,y) so that

\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0

<em>is</em> exact, which would require

\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}

\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}

Notice that

\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y

is independent of <em>x</em>, and dividing this by N(x,y)=xy gives an expression independent of <em>y</em>. If we assume \mu=\mu(x) is a function of <em>x</em> alone, then \frac{\partial\mu}{\partial y}=0, and the partial differential equation above gives

-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}

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-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}

\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x

\ln|\mu|=\ln|x|

\implies \mu=x

So, multiply the original ODE by <em>x</em> on both sides:

(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0

Now

\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy

\dfrac{\partial(x^2y)}{\partial x}=2xy

so the modified ODE is exact.

Now we look for a solution of the form F(x,y)=C, with differential

\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

The solution <em>F</em> satisfies

\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2

\dfrac{\partial F}{\partial y}=x^2y

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\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y

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So the solution to the ODE is

F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C

\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}

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