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daser333 [38]
3 years ago
7

Can someone help me solving this differential equation?

Mathematics
1 answer:
maria [59]3 years ago
8 0

Use reduction of order. Given a solution y_1(x) = x^2, look for a second solution of the form y_2(x) = y_1(x)v(x).

Compute the first two derivatives of y_2(x):

y_2 = x^2v \\\\ {y_2}' = x^2v' + 2xv \\\\ {y_2}'' = x^2v''+4xv' + 2v

Substitute them into the ODE:

x^4 (x^2v'' + 4xv' + 2v) + x^3 (x^2v' + 2xv) - 4x^2 (x^2v) = 1 \\\\ x^6v'' + 5x^5v' = 1

Now substitute w(x) = v'(x) and you end up with a linear ODE:

x^6w'+5x^5w=1

Multiply through both sides by \frac1x (if you're familiar with the integrating factor method, this is it):

x^5w'+5x^4w = \dfrac1x

Bear in mind that in order to do this, we require x\neq0. Just to avoid having to deal with absolute values later, let's further assume x>0.

Notice that the left side is the derivative of a product,

\left(x^5w\right)' = \dfrac1x

Integrate both sides with respect to x :

x^5w = \displaystyle \int\frac{\mathrm dx}x \\\\ x^5w = \ln(x) + C_1

Solve for w(x) :

w = \dfrac{\ln(x)+C_1}{x^5}

Solve for v(x) by integrating both sides:

v = \displaystyle \int \frac{\ln(x)+C_1}{x^5} \,\mathrm dx

Integrate by parts:

\displaystyle f = \ln(x) + C_1 \implies \mathrm df = \frac{\mathrm dx}x \\\\ \mathrm dg = \frac{\mathrm dx}{x^5} \implies g = -\frac1{4x^4} \\\\ \implies v = -\frac{\ln(x)+C_1}{4x^4} + \frac14 \int \frac{\mathrm dx}{x^5} \\\\ v = -\frac{\ln(x)+C_1}{4x^4} - \frac1{16x^4} + C_2 \\\\ v = -\frac{4\ln(x)+C_1}{16x^4}+C_2

Solve for y_2(x) :

\displaystyle \frac{y_2}{x^2} = -\frac{4\ln(x)+C_1}{16x^4}+C_2 \\\\ y_2 = -\frac{4\ln(x)+C_1}{16x^2} + C_2x^2

But since y_1(x)=x^2 is already accounted for, the second solution is just

\displaystyle y_2 = -\frac{4\ln(x)+C_1}{16x^2}

Still, the general solution would be

\displaystyle \boxed{y(x) = -\frac{4\ln(x)+C_1}{16x^2} + C_2x^2}

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Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp
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Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If \phi: G \longrightarrow G is an homomorphism, then it must hold

that b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1},

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)

which tells us that \phi is a homorphism. The kernel of \phi

is the set of elements g in G such that g^{n}=1. However,

\phi is not necessarily 1-1 or onto, if G=\mathbb{Z}_6 and

n=3, we have

kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}

(c) If z_1,z_2 \in \mathbb{C}^{\times} remeber that

|z_1 \cdot z_2|=|z_1|\cdot|z_2|, which tells us that \phi is a

homomorphism. In this case

kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}, if we write a

complex number as z=x+iy, then |z|=x^2+y^2, which tells

us that kern(\phi) is the unit circle. Moreover, since

kern(\phi) \neq \{1\} the mapping is not 1-1, also if we take a negative

real number, it is not in the image of \phi, which tells us that

\phi is not surjective.

(d) Remember that e^{ix}=\cos(x)+i\sin(x), using this, it holds that

\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)

which tells us that \phi is a homomorphism. By computing we see

that  kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \} and

Im(\phi) is the unit circle, hence \phi is neither injective nor

surjective.

7 0
3 years ago
I need help im going to post later a 3 more times 10 points
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Answer:

X=9  Y= 3

Step-by-step explanation:

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Wich decimal is equivalent to 122/11
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122/11= 11.09 or 11.1 depending on how you like to round numbers
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3 years ago
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A chef is going to use a mixture of two brands of Italian dressing. The first brand contains 6% vinegar, and the second brand co
rewona [7]

Answer: To make a 310mL mixture that contains 8% vinegar, use 186mL of the first brand and 124mL of the second brand.

Step-by-step explanation:We want the total mixture to be 310mL so: F + S = 310

 

Also, you want the total mixture to contain 8% vinegar so: .06F + .11S = .08(310)

 

Now, we can solve the system of equations using substitution.

 

F + S = 310

.06F + .11S = .08(310)

 

F = 310 - S      [solve 1st equation for F]

.06(310 - S) + .11S = .08(310)     [substitute into 2nd equation]

18.6 -.06S +.11S = 24.8

.05S = 6.2

S = 124

 

F + 124 = 310     [substitute answer for S back into original equation]

F = 186

 

6 0
2 years ago
Use this system of equations to answer the questions that what number would you multiply the second equation
KatRina [158]

Complete question :

Use this system of equations to answer the questions that what number would you multiply the second equation by in order to eliminate the x-terms when adding to the first equation?

What number would you multiply the second equation by in order to eliminate the y-terms when adding to the first equation?

4x – 9y = 7

–2x + 3y = 4

Answer:

2 ; 3

Step-by-step explanation:

Given the equations :

4x – 9y = 7 - - - - - (1)

–2x + 3y = 4 - - - (2)

The number to multiply equation (2) by in other to eliminate the x-term :

We multiply equation (2) by 2 ; thus we have ;

4x - 9y = 7 - - - - (1)

-4x + 6y = 8 - - - - (2) ; adding (1) and (2)

___________

____ - 3y = 15

___________

The number to multiply equation (2) by in other to eliminate the y-term :

We multiply equation (2) by 3 ; thus we have ;

4x - 9y = 7 - - - - (1)

-6x + 9y = 12 - - - - (2) ; adding (1) and (2)

___________

-2x = 19

___________

4 0
2 years ago
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