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tino4ka555 [31]

2 years ago

15

Can someone help me

1 answer:

julsineya [31]2 years ago

3 0

Answer:

ok sooo just go over the grfat and mm count it up to the dot.

Step-by-step explanation:

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Help me with my math for brainiest:)

MissTica

Answers:
1. P=3

Step by step:
P-5=-2 -move the constant to the right hand side and change its sign

P=-2+5 - calculate the sum

P=3

2. W=-2.8

Step by step:
W+13.2=10.4 - move the constant tot he right hand side and change its sign

W=10.4-13.3 - calculate the difference

W=-2.8

3. X=2/3

Step by step:
X-5/6=-1/6 - move the constant to the rig hand side and change its sign

X=-1/6+5/6 - calculate the sum

X=2/3

7 0

2 years ago

Please don give me the full answer just enough that I'm able to get it right

Zielflug [23.3K]

Answer:

190

Step-by-step explanation:

5 0

2 years ago

Solve for x. mx2 + y e = b

m_a_m_a [10]

X=(b/2m)*y-e is the answer I believe.

7 0

2 years ago

Read 2 more answers

Consider these four statements about a line that passes through two points on a plane.

Brilliant_brown [7]

I think is statement B

8 0

3 years ago

Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the

kkurt [141]

Answer:

Therefore the particular solution of the given differential equation is

$Y(x)=\frac{1}{6} x^2 ln x$

Step-by-step explanation:

The given ordinary differential equation is

$x^2y"-3xy' +4y=11x^2$

If y₁(x) =x² is a solution of ODE then it will be satisfy the ODE

y₁'(x)= 2x and y₁"(x)=2

Putting the value of y₁'(x) and y₁"(x) in x²y"-3xy'+4y=0 we get

x².2-3x.2x+4x²= 2x²-6x²+4x²=0

Therefore y₁(x) is a solution of the given differential equation.

Again,

y₂(x) =(x²ln x ) is a solution of ODE then it will be satisfy the ODE.

$y'_2=2x ln x+ x^2\times\frac{1}{x}$

$= 2x ln x+x$

$y"_2(x)=2x\times \frac{1}{x} +2lnx+1$ = 3+2 ln x

Putting the value of y₂'(x) and y₂"(x) in x²y"-3xy'+4y=0 we get

$x^2 (3+2 ln x)-3x( 2x ln \ x+x)+4x^2\ ln\ x$ =0

Therefore y₂(x) is a solution of the given differential equation.

The wronskian of y₁(x) and y₂(x) is

$W(y_1,y_2)(x)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|$

$=\left|\begin{array}{cc}x^2&x^2 ln x\\2x&2x ln x+x\end{array}\right|$

=x²(2x ln x+x)-x²ln x(2x)

=2x³ ln x +x³ - 2x³ln x

=x³≠0

Here $g(x)=\frac{y_2(x)}{y_2(x)} = ln x$

The particular solution is

$Y(x)=- y_1(x)\int\frac{y_2(x).g(x)}{W(y_1,y_2)(x)}dx + y_2(x)\int\frac{y_1(x).g(x)}{W(y_1,y_2)(x)}dx$

$=-x^2\int\frac{x^2 ln x . ln x}{x^3} dx+x^2ln x\int \frac{x^2ln x}{x^3} dx$

$=-x^2\int \frac{(lnx)^2}{x} dx+x^2 ln x\int \frac{ln x}{x} dx$

$=-x^2\int u^2 du+x^2ln x\int u dx$ Let ln x =u $\Rightarrow \frac{1}{x} dx=du$

$=-x^2 \frac{u^3}{3} +x^2 ln x \frac{u^2}{2}$

$=-\frac{x^2(lnx)^3}{3} +\frac{x^2(lnx)^3}{2}$

$=\frac{1}{6}x^2 (ln x)^3$

Therefore the particular solution of the given differential equation is

$Y(x)=\frac{1}{6} x^2 ln x$

8 0

2 years ago