Solving differential equation(separable)

In summary: If y(0) > 1 you must take y - 1 = y - 1 and you must take the negative branch of the solution. In summary, to solve the given differential equation with the given initial condition, we first separate the variables and then solve for y using the natural logarithm. However, depending on the value of y(0), we must take a different branch of the solution. If y(0) < 1, we must take the negative branch of the solution, giving y = -1. If y(0) > 1, we must take the positive branch of the solution, giving y = 3.
  • #1
aerograce
64
1
Let y be a solution of the di®erential equation,
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{y-1}{x+1}[/itex]

such that

y(0)=0;

Then y(1) is:

My attempt:

[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx

ln|y-1|= ln |x+1| +C

Problems start to occur from here:

My method:

When x=0, y=0. Hence ln0=ln0+C C=0;
When x=1, ln|y-1|= ln2 + 0;
Hence |y-1| = 2
y=3 or y=-1.

My teacher's method
[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx

ln|y-1|= ln |x+1| +C

(y-1)=A(x+1)

Because y(0)=0 Hence A=-1

And when x=1, y-1=-2 y=-1



I am quite confused. My method seems to be legit also, but why I got different answer.
 
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  • #2
aerograce said:
Let y be a solution of the di®erential equation,
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{y-1}{x+1}[/itex]

such that

y(0)=0;

Then y(1) is:

My attempt:

[itex]\frac{1}{y-1}[/itex]dy=[itex]\frac{1}{x+1}[/itex]dx

ln|y-1|= ln |x+1| +C

Problems start to occur from here:

My method:

When x=0, y=0. Hence ln0=ln0+C C=0;

You of course mean [itex]\ln 1 = \ln 1 + C[/itex] so that [itex]0 = 0 + C[/itex].

When x=1, ln|y-1|= ln2 + 0;
Hence |y-1| = 2
y=3 or y=-1.

If you look at the ODE you'll see that if [itex]y < 1[/itex] then [itex]dy/dx < 0[/itex] and if [itex]y > 1[/itex] then [itex]dy/dx > 0[/itex] (in both cases this assumes [itex]x \geq 0[/itex]). Thus if [itex]y(0) < 1[/itex] then [itex]y(x) < 1[/itex] for all [itex]x \geq 0[/itex], so you must take [itex]|y - 1| = 1 - y[/itex].
 

1. What is a differential equation?

A differential equation is an equation that relates a function or a set of functions to its derivatives. It is used to describe the relationship between a quantity and its rate of change.

2. How do you solve a separable differential equation?

To solve a separable differential equation, you need to separate the variables on either side of the equation and then integrate both sides. This will result in an equation with only one variable on each side, which can then be solved using algebraic techniques.

3. What is the meaning of the constant of integration in a separable differential equation?

The constant of integration is a constant term that is added to the solution of a differential equation. It arises due to the indefinite nature of integration and represents all possible solutions to the differential equation.

4. What is the difference between an ordinary and a partial differential equation?

An ordinary differential equation involves only one independent variable, while a partial differential equation involves more than one independent variable. This means that the solution to a partial differential equation will be a function of multiple variables.

5. Can separable differential equations be applied to real-world problems?

Yes, separable differential equations can be used to model a variety of physical phenomena in fields such as physics, engineering, and economics. Some common applications include population growth, radioactive decay, and chemical reactions.

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