Solving a differential equation with intial conditions,

In summary, the conversation is about a student who made a mistake while solving a differential equation and is seeking help to identify where they went wrong. They rearranged the equation and found the integrating factor, but did not correctly apply it to both sides of the equation. Eventually, they discovered their mistake and found the correct solution.
  • #1
skyturnred
118
0

Homework Statement



Here is the original thing:

(x[itex]^{2}[/itex]+1)y'+4x(y-1)=0, y(0)=4

Homework Equations





The Attempt at a Solution



I thought I knew the procedure.. but I got it wrong. Can someone let me know where I went wrong?

First I rearrange the equation to get the following in the form y'+p(x)y=g(x)

y'+([itex]\frac{4x}{x^{2}+1}[/itex])y=[itex]\frac{4x}{x^{2}+1}[/itex]

So I get the integrating factor to be

I(x)=e[itex]^{\int\frac{4x}{x^{2}+1}}[/itex]

That integral comes out to be 2ln(x[itex]^{2}[/itex]+1), which when raised to e, the integrating factor simply becomes

I(x)=(x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]

So the new differential equation is:

d/dx (x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]y=[itex]\frac{4x}{x^{2}+1}[/itex]

integrating both sides gets me

(x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]y=2ln(x[itex]^{2}[/itex]+1)+C

solving for y:

y=[itex]\frac{2ln(x^{2}+1)}{(x^{2}+1)^{2}}[/itex]+[itex]\frac{C}{(x^{2}+1)^{2}}[/itex]

Taking the intial conditions into account I get C=4. When I input this equation into my assignment online, it says I'm wrong.

Please help, I thought I knew this process pretty well..
 
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  • #2
You forgot to multiply the righthand side by the integrating factor.
 
  • #3
skyturnred said:

Homework Statement



Here is the original thing:

(x[itex]^{2}[/itex]+1)y'+4x(y-1)=0, y(0)=4

Homework Equations





The Attempt at a Solution



I thought I knew the procedure.. but I got it wrong. Can someone let me know where I went wrong?

First I rearrange the equation to get the following in the form y'+p(x)y=g(x)

y'+([itex]\frac{4x}{x^{2}+1}[/itex])y=[itex]\frac{4x}{x^{2}+1}[/itex]

So I get the integrating factor to be

I(x)=e[itex]^{\int\frac{4x}{x^{2}+1}}[/itex]

That integral comes out to be 2ln(x[itex]^{2}[/itex]+1), which when raised to e, the integrating factor simply becomes

I(x)=(x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]
So you are going to multiply both sides of the equation by that?

So the new differential equation is:

d/dx (x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]y=[itex]\frac{4x}{x^{2}+1}[/itex]
Did you not multiply the right side by [itex](x^2+ 1)^2[/itex]?

integrating both sides gets me

(x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]y=2ln(x[itex]^{2}[/itex]+1)+C

solving for y:

y=[itex]\frac{2ln(x^{2}+1)}{(x^{2}+1)^{2}}[/itex]+[itex]\frac{C}{(x^{2}+1)^{2}}[/itex]

Taking the intial conditions into account I get C=4. When I input this equation into my assignment online, it says I'm wrong.

Please help, I thought I knew this process pretty well..
 

FAQ: Solving a differential equation with intial conditions,

What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It is used to describe the relationship between a quantity and its rate of change over time.

Why do we need to solve differential equations?

Differential equations are used to model real-life situations in science and engineering. By solving them, we can predict the behavior of systems and make informed decisions.

What are initial conditions in a differential equation?

Initial conditions are the starting values of a function and its derivatives at a specific point. They are used to determine a unique solution to a differential equation.

How do we solve a differential equation with initial conditions?

To solve a differential equation with initial conditions, we use various methods such as separation of variables, integrating factors, and substitution. The initial conditions are used to find the constants of integration and determine the particular solution.

Can all differential equations be solved with initial conditions?

No, not all differential equations can be solved analytically with initial conditions. Some equations have no closed-form solution and require numerical methods or approximations to find a solution.

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