Differential equation/Integration problem

In summary, the conversation discusses a problem involving the equation (1) where B is a constant. The speaker tried to solve the problem by separating the variables and using Wolfram Alpha to integrate the LHS, but the solution did not work. They also attempted to solve the integral from first principles and looked for it in standard integral tables, but were unsuccessful. The speaker then asks for a solution method or clarification on whether Wolfram's solution is correct. The response suggests using a series solution for ## \int\frac{\mathrm{sinh^{-1}}\left(y\right)}{y}dy ## and provides a more general solution for ## \int\frac{\mathrm{sinh^{-1}}\left(y\right)}
  • #1
JulieK
50
0
I have the following problem

\begin{equation}
\frac{\mathrm{arcsinh}\left(y\right)}{y}\frac{dy}{dx}=B\end{equation}where B is constant. To solve the problem I separated the
variables and obtained

\begin{equation}
\int\frac{\mathrm{arcsinh}\left(y\right)}{y}dy=B \int dx\end{equation}I used Wolfram alpha to integrate the LHS and obtained an expression
which did not work for some reason. To check this I tried to do this
from first principles but the attempts led to dead end. I also could
not find such an integral in standard integral tables. Can someone
suggest a solution method to the problem or show me how to integrate
the LHS from first principles or prove Wolfram is right or wrong.
 
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  • #2
Apparently the solution for ## \int\frac{\mathrm{sinh^{-1}}\left(y\right)}{y}dy ## is a series.

For y2 < 1, the solution is (Integral 731.1 from H. B. Dwight, Tables of Integrals and Other Mathematical Data, Macmillan, NY, NY 1961.

##y - \frac{1}{2\cdot 3\cdot{3}} y^3 + \frac{1\cdot{3}}{2\cdot{4}\cdot{5}\cdot{5}} y^5 - \frac{1\cdot{3}\cdot{5}}{2\cdot{4}\cdot{6}\cdot{7}\cdot{7}} y^7 + . . .##

for a more general solution, let y = x/a, and for y > 1 or < 1 the solution is somewhat different with even exponents.
## \int\frac{\mathrm{sinh^{-1}}\left(y\right)}{y^m}dy ## can be solved analytical for some m, at least for m = 2, 3
 
  • Like
Likes JulieK
  • #3
Thanks Astronuc for your useful remark!
I solved the problem by testing Wolfram numerically using numerical integration. I noticed that Wolfram expression produces large errors in some cases.
Replacing Wolfram expression with numerical integration I obtained almost perfect results.
 

FAQ: Differential equation/Integration problem

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model a variety of physical phenomena, including motion, growth, and decay.

What is integration?

Integration is a mathematical operation that is the inverse of differentiation. It involves calculating the area under a curve and is used to solve a wide range of problems in mathematics, science, and engineering.

How do you solve a differential equation?

There is no one method for solving a differential equation, as it depends on the type and complexity of the equation. Some common techniques include separation of variables, substitution, and using special functions. In some cases, numerical methods may also be used.

What are the applications of differential equations?

Differential equations have numerous applications in various fields, including physics, chemistry, biology, economics, and engineering. They are used to model and understand dynamic systems, such as the behavior of particles, population growth, and electrical circuits.

How is integration used in real life?

Integration has many real-life applications, such as calculating the area under a curve to find the volume of a shape, determining the displacement of an object from its velocity, and finding the average value of a function. It is also used in economics to calculate consumer surplus and in statistics to calculate probabilities.

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