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3 years ago

10(6x-6y)+6(5x+5y) What is the answer

Mathematics

2 answers:

mina [271]3 years ago

6 0

Answer:

I think it 90x-30y hopefully it helps

solniwko [45]3 years ago

4 0

Answer:

solution=

10( 6x - 6y) + 6(5x + 5y)

=(60x - 60y) + (30x + 30y)

= 60x + 30x -60y + 30y

=90x - 30y

=30( 60x -y)

Step-by-step explanation:

At first, we multiply the 10 with (6x -6y) & 6 with ( 5x + 5y) after that we gets some result and we have to open the small brackets and then we have to make corrects its position of same component then add or subtract wthever there is at last there is 30 common we have put it out and remains other.

Hope This will help yUh!

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Solve this problem by using variation of parameters method. y''-y=coshx.

Gekata [30.6K]

$y''-y=0\implies r^2-1=0\implies r=\pm1$
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For the nonhomogeneous ODE

$y''-y=\underbrace{\cosh x}_{f(x)}$

we're looking for a particular solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the two fundamental solutions.

We have

$W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1$

so we're left with

$u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x$
$u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x$

so that the particular solution is

$y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x$
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As $y_1$ already accounts for the $\cosh x$ term in $y_p$ , we're left with the general solution

$y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x$