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Colt1911 [192]
3 years ago
10

Consider the differential equation

B2%7D%20%2B6xy" id="TexFormula1" title=" x^{2} \frac{dy}{dx} =6 y^{2} +6xy" alt=" x^{2} \frac{dy}{dx} =6 y^{2} +6xy" align="absmiddle" class="latex-formula"> which may be considered either as a homogenous equation or as a Bernoulli equation.
If we make the substitution y(x)=xv(x) relevant to homogenous equations, we obtain \frac{dv}{dx}=
If we make the substitution z(x)= (y(x))^{-1} relevant to homogenous equations, we obtain \frac{dz}{dx}=
Using either (or both) of these methods, solve the initial value problem for the above equation where y(3)=6. Find the interval of validity of this solution.
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0
As a Bernoulli equation:

x^2\dfrac{\mathrm dy}{\mathrm dx}=6y^2+6xy\iff x^2y^{-2}\dfrac{\mathrm dy}{\mathrm dx}-6xy^{-1}=6

Let z=y^{-1}\implies\dfrac{\mathrm dz}{\mathrm dx}=-y^{-2}\dfrac{\mathrm dy}{\mathrm dx}. The ODE becomes

-x^2\dfrac{\mathrm dz}{\mathrm dx}-6xz=6
x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-6x^4
\dfrac{\mathrm d}{\mathrm dx}[x^6z]=-6x^4
x^6z=-6\displaystyle\int x^4\,\mathrm dx
x^6z=-\dfrac65x^5+C
z=-\dfrac6{5x}+\dfrac C{x^6}
y^{-1}=-\dfrac6{5x}+\dfrac C{x^6}
y=\dfrac1{\frac C{x^6}-\frac6{5x}}
y=\dfrac{5x^6}{C-6x^5}

With y(3)=6, we get

6=\dfrac{5(3)^6}{C-6(3)^5}\implies C=\dfrac{4131}2

so the solution is

y=\dfrac{5x^6}{\frac{4131}2-6x^5}=\dfrac{10x^6}{4131-12x^5}

which is valid as long as the denominator is not zero, which is the case for all x\neq\sqrt[5]{\dfrac{4131}{12}}.
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4 0
2 years ago
This is a System Of Equations and I'd like it to get solved by the Elimination or Substitution Method.x+2y=83x+5y=8
Sonbull [250]

Answer:

x=133 y=-25

Step-by-step explanation:

I'll do both ways for you. So let's start with Substitution:

With the sub method, you have to set both equations equal to each other by setting them equal to the same variable. Since there is no coefficient in front of both x's in both equations, that variable will be easiest to solve for.

x + 2y = 83    &    x + 5y = 8

Solve for x.

x = 83 - 2y     &    x = 8 - 5y

Once you have solved for x, set each equation equal to one another and solve for y now.

83 - 2y = 8 - 5y

Isolate all variables to one side:

83 = 8 - 3y

Now subtract the 8 to fully isolate the y variable:

75 = -3y

Divide by -3:

-25 = y   Now that you have your first variable, plug it into one of the original equations and solve for x.

x + 2(-25) = 83

x - 50 = 83

x = 133

Now for the Elimination method. For this method you need to get rid of a variable by either subtracting/adding each equation together. Again, since you can subtract and x from both equations, you will be left with only the y variable to solve:

Put each equation on top of one another and subtract:

  x + 2y = 83

- (x + 5y = 8)

The x's will cancel out:

(x - x) + (2y - 5y) = (83 - 8)

Simplify:

-3y = 75

Solve for y

y = -25

Then, plug y = -25 into one of the original equations:

x + 5(-25) = 8

Solve for x:

x - 125 = 8

x = 133

Hope this helps!

8 0
2 years ago
Given ​f(x)=20x+11​, find ​f(4​)
kakasveta [241]

Answer:

91

Step-by-step explanation:

Substitute 4 in place of x

f(4)=20*4+11

f(4)=80+11

f(4)=91

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