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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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Consider the figure below. Line FG is parallel to line JK.
Svet_ta [14]

Answer:

A. a = 10

B. m<FNT = 120°

C. m<KTU = 60°

Step-by-step explanation:

A. (7a + 50)° and (14a - 20)° are corresponding angles. Therefore:

(7a + 50)° = (14a - 20)°

Use this equation to find the value of a

7a + 50 = 14a - 20

Combine like terms

7a - 14a = - 50 - 20

-7a = -70

Divide both sides by -7

-7a/-7 = -70/-7

a = 10

B. m<FNT = (14a - 20)° (alternate interior angles are congruent)

Plug in the value of a

m<FNT = 14(10) - 20

m<FNT = 140 - 20

m<FNT = 120°

C. m<KTU + (14a - 20)° = 180° (linear pair)

Plug in the value of a

m<KTU + 14(10) - 20 = 180

m<KTU + 120 = 180

m<KTU + 120 - 120 = 180 - 120

m<KTU = 60°

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