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google it

Oh Thanks.

Also:

Real number: \Bbb{R}  \(\Bbb{R}\)

Rational number: \Bbb{Q}   \(\Bbb{Q}\)

Integers: \Bbb{Z}   \(\Bbb{Z}\)

Natural number: \Bbb{N}   \(\Bbb{N}\)

Complex number: \Bbb{C} \(\Bbb{C}\)

\(0.25x+3=0.1x+4.5 \)

\(25x+300=10x+450\)

\(15x=150\)

\(x=10\)

Solve the linear equation x ( dy(x))/( dx) - (x + 1) y(x) = -e^x (x^2 + 1):
Rewrite the equation:
( dy(x))/( dx) + ((-x - 1) y(x))/x = -(e^x (x^2 + 1))/x
Let μ(x) = e^( integral(-x - 1)/x dx) = e^(-x)/x

Multiply both sides by μ(x):
(e^(-x) ( dy(x))/( dx))/x + ((e^(-x) (-x - 1)) y(x))/x^2 = -(x^2 + 1)/x^2
Substitute (e^(-x) (-x - 1))/x^2 = ( d)/( dx)(e^(-x)/x):
(e^(-x) ( dy(x))/( dx))/x + ( d)/( dx)(e^(-x)/x) y(x) = -(x^2 + 1)/x^2
Apply the reverse product rule g ( df)/( dx) + f ( dg)/( dx) = ( d)/( dx)(f g) to the left-hand side:

( d)/( dx)((e^(-x) y(x))/x) = -(x^2 + 1)/x^2
Integrate both sides with respect to x:
 integral( d)/( dx)((e^(-x) y(x))/x) dx = integral-(x^2 + 1)/x^2 dx

Evaluate the integrals:
(e^(-x) y(x))/x = -x + 1/x + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(x) = e^(-x)/x:
Answer: |y(x) = e^x x (-x + 1/x + C1)

Solve the linear equation (y(x))/sqrt(x^2 + 1) + ( dy(x))/( dx) = 1:
Let μ(x) = e^( integral1/sqrt(x^2 + 1) dx) = e^(sinh^(-1)(x)).

Multiply both sides by μ(x):
e^(sinh^(-1)(x)) ( dy(x))/( dx) + (e^(sinh^(-1)(x)) y(x))/sqrt(x^2 + 1) = e^(sinh^(-1)(x))

Substitute e^(sinh^(-1)(x))/sqrt(x^2 + 1) = ( d)/( dx)(e^(sinh^(-1)(x))):
e^(sinh^(-1)(x)) ( dy(x))/( dx) + ( d)/( dx)(e^(sinh^(-1)(x))) y(x) = e^(sinh^(-1)(x))

Apply the reverse product rule g ( df)/( dx) + f ( dg)/( dx) = ( d)/( dx)(f g) to the left-hand side:
( d)/( dx)(e^(sinh^(-1)(x)) y(x)) = e^(sinh^(-1)(x))

Integrate both sides with respect to x:
 integral( d)/( dx)(e^(sinh^(-1)(x)) y(x)) dx = integral e^(sinh^(-1)(x)) dx

Evaluate the integrals:
e^(sinh^(-1)(x)) y(x) = 1/2 (sinh^(-1)(x) + x (x + sqrt(x^2 + 1))) + c_1, where c_1 is an arbitrary constant.

Divide both sides by μ(x) = e^(sinh^(-1)(x)):
Answer: |y(x) = (1/2 (sinh^(-1)(x) + x (x + sqrt(x^2 + 1))) + C1)/e^(sinh^(-1)(x))

y'+6y/(x+2)=1/(x+2)²   multiply both sides by (x+ 2)^6

y' (x + 2)^6  +  6y(x + 2)^5  =  (x + 2)^4

Let u   = y (x + 2)^6    →  du =   y' (x + 2)^6  +  6y(x + 2)^5  

So we have

du   = (x + 2)^4    integrate both sides

∫ du   = ∫  (x + 2)^4 dx

u =  (x + 2)^5 / 5    +  C₁

y (x + 2)^6  =  (x + 2)^5 / 5    +  C₁        divide through by   ( x + 2)^6

y  =   1 / [5( x+ 2)]  +  C₁ / ( x + 2)^6

y  =   1 / [5x + 10]  +  C₁ / ( x + 2)^6

Oh thank you.

I am not very careful

xy'+2y=x²-3        multiply both sides by x

x^2y'+2xy=x^3-3x

Let u  =  x^2*y   →  du   =  x^2y'  + 2xy

So we have

du  = x^3-3x      integrate both sides

∫ du   = ∫ x^3 -3x dx

u  = x^4/4  - (3/2)x^2  + C₁

x^2*y = x^4/4  - (3/2)x^2  + C₁     divide through by   x^2

y  =  x^2/4  - (3/2)  +  C₁ / x^2

It went wrong at the end Max.

It's better if the constant is on the x side, and then

\(\displaystyle \ln \frac{1}{1+x^{2}}+\ln(K)=\ln(y)\),

\(\displaystyle \ln\frac{K}{1+x^{2}}=\ln(y)\),

\(\displaystyle y = \frac{K}{1+x^{2}}\).

Tiggsy

y'+y* tanx=1/cosx

y' + y [tan x]  = sec x          multiply both sides by sec x

y'sec x  + y [  sec x tan x ]   =  sec^2 x

Let u  =  y sec x   →  du   =  y'sec x  + y*  sec x tan x

So

du  = sec^2 x     integrate both sides

∫ du   = ∫ sec^2 dx

u =  tan x + C₁

y sec x   =  tan x + C₁   

y [ 1/cos x ]   = sin x / cos x  + C₁      multiply through by cos x

y =   sin x + C₁ cos x

Sorry, missed a dt at the end of the last equation.

If t = tan(x/2), then tan(x) = 2t/(1 - t^2) and, from the rt-angled triangle, sin(x) = 2t/(1 + t^2) and cos(x) = (1 - t^2)/(1 + t^2) .

dt = (1/2)sec^2(x/2)dx, so dx = 2/(1 + t^2)dt.

On substitution, the (1 + t^2) terms tend to cancel.

If t = tan(x/2), then tan(x) = 2t/(1 - t^2) and, from the rt-angled triangle, sin(x) = 2t/(1 + t^2) and cos(x) = (1 - t^2)/(1 + t^2) .

dt = (1/2)sec^2(x/2)dx, so dx = 2/(1 + t^2).

On substitution, the (1 + t^2) terms tend to cancel.

OK. I thought it was correct...... Let me do it myself......

\(2xy\text{ dx} + (1+x^2) \text{ dy} = 0\\ 2xy\text{ dx} = -(1+x^2) \text{ dy}\\ -\dfrac{2x}{1+x^2} \text{ dx} = \dfrac{\text{ dy}}{y}\\ \displaystyle\int -\dfrac{2x}{1+x^2} \text{ dx} = \displaystyle\int\dfrac{\text{ dy}}{y}\\ \text{ Let }u = 1 + x^2,\text{ du} = 2x \text{ dx}\\ -\displaystyle\int \dfrac{\text{ du}}{u} = \displaystyle\int\dfrac{\text{ dy}}{y}\\ -\ln |u| = \ln |y| + C\\ \ln |\dfrac{1}{u}| = \ln |y| + C\\ \ln |\dfrac{1}{1+x^2}|= \ln |y| + C\\ \text{ Set }C = \ln K\\ \ln (\dfrac{1}{1+x^2})=\ln |y + K|\\ |y + K| = \dfrac{1}{1+x^2}\\ y + K = \pm \dfrac{1}{1+x^2}\\ y = K\pm\dfrac{1}{1+x^2}\)

Thank you Max,

It looks like it might be useful but I will have to take a better look before I really know :)

I looked up the Internet and this may be useful.

Write an equation of the ellipse with focus at (0,+-9) and vertices at (0,+-13)

  [ The major axis is along  the  y  axis]

We need to find " b"  = √[a^2  − c^2] =  √ [ 13^2 − 9^2]  = √ [169 − 81]  = √88

So.... the equation is

y^2 / a^2     +     x^2 / b^2      =   1

y^2 / 169   +     x^2 / 88      =   1

Here's  the graph :   https://www.desmos.com/calculator/wsiick7axh

Hi, thanks for repsonding to my answer, we always like askers to respond because then we are certain we know I answers have been read, considered and hopefully tearned from. :))

I do not think I talked about "a system of negative numbers"  There is no such thing......

I think that you cannot quite see the relevance of everything I have said, and that is fine, I see the relevance but you do not really need to. :)

YES you do need to understand how numbers work and why different calculators give different answers.  It is always up to the user to get the best out of his machines :)

As far as the authors statement goes, I will repeat:

"He means it is easy to see because i^2=-1 but he is wrong because he has not used brackets properly.

By not using brackets properly he has not followed proper convention.

He has overlooked this because, in the context of what he was talking, his meaning was clear, he thought so anyway :)"

Hi Majid,

I am not going to attempt your calculus but it is great to finally see you posting on the forum  :D

I always like looking at these answers too if anyone want to have a go. :)

When will 3.00+.25x and 4.50+.10x cast the same amount of money

Well they have to be equal don't they:)

\(3.00+.25x = 4.50+.10x\)

 subtract 0.10x from BOTH sides

\(3.00+0.15x = 4.50\)

NOW subtract 3.00 from both sides and then think about how to finish it :)

Hi guest,

Could you (or someone else) please elaborate on the first step.  Thanks :)

Hi Helpls, welcome to our web2 forum :)

what is the derivative of 1+0.00003t+0.0000004t^2? with steps please

The derivative of 1 is 0

The derivative of  0.00003t

\(\frac{d}{dx}\;0.00003t^1\\ =1*0.00003t^{1-1}\\ =0.00003t^{0}\\ =0.00003*1\\ =0.00003\\\)

and

The derivative of  0.0000004t^2 is  

\(=2* 0.0000004t^{2-1}  \\ = 0.0000008t  \\\)

So

\(\frac{d}{dx}\;[1+0.00003t+0.0000004t^2] =0.00003+0.0000008t\\\)

Thank you for your reply.

Just to be sure I understand, to the first question, your example says that the system of negative numbers provides -2 as a square root of 4.  While the other square root of 4 is 2 (which is not from the system of negative numbers), there is no reason to think the two roots are equal to each other. Similarly, as one cube root of 8 is from the system of imaginary numbers, there is no reason to think it is equal to 2, which is also a cube root of 8 but is not from the system of imaginary numbers.  In sum, roots of the same number from different systems of numbers should not be considered equal to each other. 

As to the other question, I still find it awkward that the author says it is 'easy' to see, I guess because typically a teacher or lecturer who uses this terminology I think  uses it to explain the outcome of a a process of some kind (add, subtract, multiply, etc) but here negative 1 is not the outcome of a process but is according to a basic definition that i squared = negative 1..   

Also, I'm glad you gave examples of how the calculator goes about its work because in the past I have tried to use the calculator to help understand basic issues like this, but now I see my own $20 calculator, depending on how the format of the keys are pressed, in this example of i squared, returns +1, -1, or Domain error.  So, to avoid misinterpreting the result, I really must understand the subject thoroughly when using it.

If you use 3.14 for pi.....your answer is good......

0.1% of 100€

\(100\ \$\times0.1\%=\frac{100\ \$\times 0.1}{100}=0.10\ \$ \) 

  !

Would it not be 28.26? I needed to check my work and 3.14 x 9 would be 28.26.