Melissa's question at Yahoo Answers regarding solving a linear first order ODE

In summary, we are given a first order linear ODE and we solve it by using the integrating factor method and integration by parts. The general solution is given by $\displaystyle f(y)=\frac{\sin(2y)-2\cos(2y)}{5}+c_1e^{-y}$.
  • #1
MarkFL
Gold Member
MHB
13,288
12
Here is the question:

Solve this differential Equation: df/dy(t) + f(y) =sin(2y)?

df/dy(t) + f(y) =sin(2y)

I'm really stuck on how to start this differential equation. Thanks!

Here is a link to the original question:

Solve this differential Equation: df/dy(t) + f(y) =sin(2y)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
  • #2
We are given to solve the first order linear ODE:

$\displaystyle \frac{df}{dy}+f(y)=\sin(2y)$

I am assuming $\displaystyle f$ is the dependent variable and $\displaystyle y$ is the independent variable, and that the (t) is a typo.

We may begin by calculating our integrating factor $\displaystyle \mu(y)$:

$\displaystyle \mu(y)=e^{\int\,dy}=e^y$

Multiply the ODE by the integrating factor:

$\displaystyle e^y\frac{df}{dy}+f(y)e^y=e^y\sin(2y)$

Now, we may rewrite the left side as the differentiation of a product:

$\displaystyle \frac{d}{dy}\left(e^yf \right)=e^y\sin(2y)$

Integrate with respect to $\displaystyle y$:

$\displaystyle \int\frac{d}{dy}\left(e^yf \right)\,dy=\int e^y\sin(2y)\,dy$

On the right side, we may use integration by parts:

$\displaystyle u=\sin(2y)\,\therefore\,du=2\cos(2y)\,dy$

$\displaystyle dv=e^y\,dy\,\therefore\,v=e^y$

and so we may state:

$\displaystyle I=\int e^y\sin(2y)\,dy=e^y\sin(2y)-2\int e^y\cos(2y)\,dy$

Now, using integration by parts again:

$\displaystyle u=\cos(2y)\,\therefore\,du=-2\sin(2y)\,dy$

$\displaystyle dv=e^y\,dy\,\therefore\,v=e^y$

and we have:

$\displaystyle I=e^y\sin(2y)-2\left(e^y\cos(2y)+2\int e^y\sin(2y)\,dy \right)$

$\displaystyle I=e^y\sin(2y)-2e^y\cos(2y)-4I$

Solve for $\displaystyle I$:

$\displaystyle I=\frac{e^y(\sin(2y)-2\cos(2y))}{5}+c_1$

Now, back to integrating the ODE, we have:

$\displaystyle e^yf=\frac{e^y(\sin(2y)-2\cos(2y))}{5}+c_1$

Hence, the general solution is given by:

$\displaystyle f(y)=\frac{\sin(2y)-2\cos(2y)}{5}+c_1e^{-y}$
 
Last edited:

FAQ: Melissa's question at Yahoo Answers regarding solving a linear first order ODE

What is a linear first order ODE?

A linear first order ODE is a type of differential equation that involves only the first derivative of the dependent variable and can be written in the form y' + p(x)y = q(x), where p(x) and q(x) are functions of x.

What is the process for solving a linear first order ODE?

The most common method for solving a linear first order ODE is called the method of integrating factors. This involves multiplying both sides of the equation by an integrating factor, which is a function that helps to cancel out the p(x) term. The resulting equation can then be solved using standard integration techniques.

What are the applications of solving linear first order ODEs?

Linear first order ODEs are used in many areas of science and engineering to model a wide variety of phenomena. They can be used to describe the rate of change of a quantity over time, such as in population growth, chemical reactions, and electrical circuits.

Can all linear first order ODEs be solved analytically?

No, not all linear first order ODEs can be solved analytically. Some equations may be too complex to solve using standard methods, and in these cases, numerical methods must be used to approximate a solution.

What are some common mistakes to avoid when solving a linear first order ODE?

Some common mistakes when solving linear first order ODEs include forgetting to multiply by the integrating factor, making arithmetic errors when integrating, and forgetting to include the constant of integration. It is important to double-check each step of the solution process to avoid mistakes.

Back
Top