Solving a Separable Variable Differential Equation Using U-Substitution

  • MHB
  • Thread starter karush
  • Start date
C = 2\tan{\sqrt{y}} + C \)Using the initial substitution, this becomes:\( = 2\tan{\sqrt{y}} + C = 2\tan{\sqrt{y}} + C = 2\tan{\sqrt{y}} + C \)In summary, the conversation discusses solving the differential equation \( \dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}} \) by separating variables and using the substitution \( u=\sqrt{y} \) to simplify the integral. The final solution is \( 2\tan{\sqrt{y}} +
  • #1
karush
Gold Member
MHB
3,269
5
Solve the de $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$
sepate variables
$$\displaystyle \dfrac{dy}{\sqrt{y}\, \cos^2{\sqrt{y}} }=\dfrac{1}{7}\, dx
\implies \int{\dfrac{{d}y}{\sqrt{y}\,\cos^2{\left(\sqrt{y} \right) }} }
= \int{ \dfrac{1}{7}\,{d}x}$$
ok i think u subst is next ... maybe...
$$u=\sqrt{y} \therefore du=\dfrac{{d}y}{2\,\sqrt{y}}$$
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
karush said:
Solve the de $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$
sepate variables
$$\displaystyle \dfrac{dy}{\sqrt{y}\, \cos^2{\sqrt{y}} }=\dfrac{1}{7}\, dx
\implies \int{\dfrac{{d}y}{\sqrt{y}\,\cos^2{\left(\sqrt{y} \right) }} }
= \int{ \dfrac{1}{7}\,{d}x}$$
ok i think u subst is next ... maybe...
$$u=\sqrt{y} \therefore du=\dfrac{{d}y}{2\,\sqrt{y}}$$
You stopped too soon! What did you get when you did the substitution?

-Dan
 
Last edited by a moderator:
  • #3
using your sub \( u = \sqrt{y} \) ...

\( \displaystyle 2 \int \dfrac{1}{\cos^2{\sqrt{y}}} \cdot \dfrac{dy}{2\sqrt{y}} = 2\int \sec^2{u} \, du \)
 

FAQ: Solving a Separable Variable Differential Equation Using U-Substitution

What is a separable variable differential equation?

A separable variable differential equation is a type of differential equation where the variables can be separated into two functions that are multiplied together. This allows for the equation to be solved by integrating both sides separately.

What is U-substitution?

U-substitution is a technique used to solve integrals by substituting a new variable, typically denoted as u, in place of a more complicated expression. This allows for the integral to be simplified and solved more easily.

How do you solve a separable variable differential equation using U-substitution?

To solve a separable variable differential equation using U-substitution, first separate the variables on either side of the equation. Then, choose a new variable, u, to substitute in for one of the variables. Next, take the derivative of u with respect to the original variable and substitute it into the equation. Finally, integrate both sides and solve for the original variable.

Are there any limitations to using U-substitution for solving differential equations?

Yes, there are some limitations to using U-substitution. It is only applicable to integrals that can be written in the form of u du, and it may not work for more complex integrals that require other techniques such as integration by parts.

Can you provide an example of solving a separable variable differential equation using U-substitution?

Yes, for example, given the differential equation dy/dx = x^2y, we can separate the variables to get dy/y = x^2dx. Then, we can substitute u = x^2 and du = 2xdx into the equation to get dy/y = du/2. Integrating both sides gives ln|y| = u/2 + C, where C is the constant of integration. Solving for y gives y = Ce^(x^2/2), which is the general solution to the differential equation.

Similar threads

Replies
3
Views
3K
Replies
4
Views
1K
Replies
5
Views
1K
Replies
6
Views
1K
Replies
6
Views
2K
Back
Top