I'm supposed to find the expression 'y', when
(x^2+100)y'=23
When y(10sqrt(3))=0
Also, I have to find the indefite integral
of
xsqrt(19x-7), using integration tables in the back of the book "Calculus of a Single Variable, 8th Edition." Tables are in appendix B
I used the formula
[2/[b(2n-3)]]*([u^n]*(a+bu)^[3/2]-na[integral of [u^(n-1)sqrt[a+bu]]]
I've used
a=-7
b=19
u=x
du=xd
But it doesn't seem to work.
Unless you really have to finish a problem in a hurry, never use integration tables or other tools for problem solving. Only use them to very the solution when you are done.
(x^2+100)y'=23 ---->
dy/dx = 23/[x^2 + 100] ----->
y = Integral of dy =
Integral of 23 dx/[x^2 + 100], looks like an arctan to me.
Integral of x sqrt(19x-7)dx ?
Write the integral in terms of functins you do know the inegral of. Rewrite the factor of x as follows:
x = 1/19 (19 x) =
1/19 (19 x - 7 + 7) =
1/19 (19 x - 7) + 7/19
I think you'll now see the answer. These sorts of tricks cannot be learned by just looking in integral tables.
I got the first one, thanks for that.
But on the second one, are you saying that x=(1/19)*(19x-7)+(7/19)?
If so, would I just multiply the entire thing out, or what?
To find the expression for 'y', we can start by rearranging the given differential equation:
(x^2 + 100)y' = 23
y' = 23 / (x^2 + 100)
Now we have a separable differential equation. We can separate the variables and integrate both sides:
∫(1/y) dy = ∫(23 / (x^2 + 100)) dx
Integrating the left side gives:
ln|y| = ∫(23 / (x^2 + 100)) dx
To find the definite integral, we need the limits of integration. We are given y(10√3) = 0, which means y = 0 when x = 10√3. Let's integrate from x = a to x = 10√3:
ln|y| = ∫(23 / (x^2 + 100)) dx from a to 10√3
To solve this integral, we need to evaluate it. However, instead of using integration tables, we can solve it directly using trigonometric substitution.
Let's substitute x = 10 tanθ. Then dx = 10 sec^2θ dθ.
We also need to express the limits of integration in terms of θ.
When x = 10√3, θ = π/3 (since tan(π/3) = √3).
Now the integral becomes:
ln|y| = ∫(23 / ((10 tanθ)^2 + 100)) (10 sec^2θ) dθ from 0 to π/3
The expression inside the integral simplifies to:
23 / (100 tan^2θ + 100) = 23 / (100 (tan^2θ + 1)) = 23 / (100 sec^2θ)
Substituting these values, the integral becomes:
ln|y| = ∫(23 / (100 sec^2θ)) (10 sec^2θ) dθ from 0 to π/3
Simplifying this expression further, we have:
ln|y| = ∫(230 / (100 cos^2θ)) dθ from 0 to π/3
Now, integrating this expression, we have:
ln|y| = [230 / 100] ∫(1 / cos^2θ) dθ from 0 to π/3
The integral of sec^2θ is tanθ, so:
ln|y| = [230 / 100] [tanθ] from 0 to π/3
Evaluating this expression gives:
ln|y| = [230 / 100] [tan(π/3) - tan(0)]
Since tan(π/3) = √3 and tan(0) = 0, we have:
ln|y| = [230 / 100] [√3 - 0] = (√3 * 230) / 100
Finally, we can exponentiate both sides to solve for 'y':
|y| = e^[(√3 * 230) / 100]
Since y = 0 at x = 10√3, we can conclude that:
y = e^[(√3 * 230) / 100] or y = -e^[(√3 * 230) / 100]
Note: Please double-check your integral formula and ensure that you have used the correct constants and variables from your problem.