Solve Diff. Eq. w/ Initial Conditions: y' + 2y = te^-2t | Homework Help

In summary, the conversation is about solving a differential equation and finding the initial conditions. The individual asking for help has shown their attempt at a solution but is unsure if it is correct. After checking for arithmetic mistakes, it is determined that their reasoning for finding the value of c is incorrect. The correct answer is y = (t^2-1)e^-2t/2.
  • #1
dwilmer
11
0

Homework Statement



Solve the differential equation: y' + 2y = te^-2t with initial conditions y(1) = 0?
Please show me what I am doing wrong... (or right).. also please don't show me shortcut i need to know where I am going wrong thanks.



Homework Equations





The Attempt at a Solution



first i multiplied whole equation by integrating factor e^2t (e^integral(2) = e^2t)
after rearranging this gave me

e^2t(y) = integ te^-2t * e^2t

i then simplified RHS to get:
e^2t(y) = integ t

then i integrated RHS to get

e^2t(y) = (t^2)/2 + c

then i isolated y to get:

y = ((t^2)e^-2t) / 2 + ce^-2t

then i applied the initial condition y(1) = 0
this gave me

0 = e^-2 / 2 + ce^-2

then i multiplied equation by 2

0 = e^-2 + ce^-2

then i factored out e^-2

0 = e^-2 (1 + c)

then i reasoned that c must = 1 and put this back into the explicit equation

y = ((t^2)e^-2t) /2 + (1) e^-2t

is this correct? the book answer says y = (t^2 -1) e^-2t /2

THANKS FOR ANY HELP
 
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  • #2


Look for a couple of arithmetic mistakes after you used the initial condition y(1)=0.
 
  • #3


dwilmer said:

Homework Statement



Solve the differential equation: y' + 2y = te^-2t with initial conditions y(1) = 0?
Please show me what I am doing wrong... (or right).. also please don't show me shortcut i need to know where I am going wrong thanks.



Homework Equations





The Attempt at a Solution



first i multiplied whole equation by integrating factor e^2t (e^integral(2) = e^2t)
after rearranging this gave me

e^2t(y) = integ te^-2t * e^2t

i then simplified RHS to get:
e^2t(y) = integ t

then i integrated RHS to get

e^2t(y) = (t^2)/2 + c

then i isolated y to get:

y = ((t^2)e^-2t) / 2 + ce^-2t

then i applied the initial condition y(1) = 0
this gave me

0 = e^-2 / 2 + ce^-2

then i multiplied equation by 2

0 = e^-2 + ce^-2

then i factored out e^-2

0 = e^-2 (1 + c)

then i reasoned that c must = 1 and put this back into the explicit equation
Then you reasoned wrong. c is not equal to 1.

y = ((t^2)e^-2t) /2 + (1) e^-2t

is this correct? the book answer says y = (t^2 -1) e^-2t /2

THANKS FOR ANY HELP
 

FAQ: Solve Diff. Eq. w/ Initial Conditions: y' + 2y = te^-2t | Homework Help

What is a differential equation?

A differential equation is a mathematical equation that relates one or more functions to their derivatives. It describes how a function changes over time or space.

What is an initial condition?

An initial condition is a specified value or set of values for the dependent variable(s) in a differential equation at the starting point of the problem. It helps to determine the unique solution for the differential equation.

How do you solve a differential equation with initial conditions?

To solve a differential equation with initial conditions, you need to first find the general solution of the equation by using various methods such as separation of variables, integrating factors, or substitution. Then, plug in the initial conditions to determine the specific solution that satisfies the given conditions.

What is the purpose of the "y' + 2y = te^-2t" in the problem?

The "y' + 2y = te^-2t" represents the differential equation itself. It describes the relationship between the function y, its derivative y', and the independent variable t. To solve the problem, we need to find the value of y that satisfies this equation.

Why is it important to include initial conditions in a differential equation problem?

Including initial conditions in a differential equation problem is important because it helps to determine the specific solution that satisfies the given conditions. Without initial conditions, the general solution of a differential equation would have many possible solutions, making it difficult to find the specific solution that meets the requirements of the problem.

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