I have a Wronskian Question?If the Wronskian W of f and g is t^2*e^t

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In summary, the Wronskian W of f and g is t^2*e^t and f(t)=t. The question is asking to find g(t) using the given information. The process involves setting up the equation tg'-t'g = t^2*e^t and using the integrating factor µ= e^-t. After simplifying, the answer should be te^t+ct. However, there seems to be a discrepancy with the term t-1, which does not belong.
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I have a Wronskian Question?
If the Wronskian W of f and g is t^2*e^t and if f(t)=t, find g(t).

I have tried setting up this problem:

tg'-t'g = t^2*e^t
tg'-g = t^2*e^t

Setting up the integrating factor, µ= e^∫-1 --> µ= e^-t
(e^-t)t*g' - (e^-t)*g = (e^-t)(t^2*e^t)

so preferably I would want to be able to set up the equation as (fg)' = (e^-t)(t^2*e^t)

but the derivative of (e^-t)t is not (e^-t).

The answer is supposed to be te^t+ct
 
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  • #2


The quotient rule is handy here
f g'-f' g=f2(g/f)'

f g'-f' g=t2(g/t)'=t2et
 
  • #3


I'm still not getting the right answer.

I get it down to (te^t)(t-1)+ct
 
  • #4


Where are you getting that t-1? It clearly does not belong.
 
  • #5
, where c is a constant.

I would approach this problem by first reviewing the definition of a Wronskian and its properties. The Wronskian is a determinant that is used to determine the linear independence of a set of functions. In this case, the Wronskian of f and g is t^2*e^t, which means that these two functions are linearly independent.

Next, I would consider the given information that f(t)=t. This means that f'(t)=1. Using this information, we can rewrite the Wronskian equation as:

tg'(t)-g(t)=t^2*e^t

We can then use the method of variation of parameters to find the particular solution for g(t). This involves assuming that g(t) can be written as g(t)=u(t)f(t), where u(t) is a function that we need to determine.

Substituting this into the Wronskian equation, we get:

t(u'(t)f(t)+u(t)f'(t))-u(t)f(t)=t^2*e^t

Simplifying and rearranging, we get:

u'(t)f(t)=t*e^t

Integrating both sides with respect to t, we get:

u(t)=te^t+C

where C is a constant of integration.

Therefore, the particular solution for g(t) is:

g(t)=u(t)f(t)=te^t*f(t)=te^t*t=te^t+ct

where c is a constant.

In conclusion, using the method of variation of parameters, we can determine that g(t) is equal to te^t+ct, where c is a constant, given the information that the Wronskian of f and g is t^2*e^t and f(t)=t.
 

What is a Wronskian?

A Wronskian is a mathematical tool used to determine if a set of functions are linearly independent. It is denoted by W and is calculated using the derivatives of the functions.

How is the Wronskian calculated?

The Wronskian is calculated by taking the determinant of a matrix formed by the derivatives of the functions. In the case of f and g, the Wronskian is calculated as W = det([f', g']).

What does it mean if the Wronskian is equal to t^2*e^t?

If the Wronskian W of f and g is equal to t^2*e^t, it means that the two functions f and g are linearly dependent. This means that one function can be written as a linear combination of the other, and therefore they do not form a basis for the vector space.

Why is the Wronskian important?

The Wronskian is important because it is used to determine the linear independence of a set of functions. It is also used in solving differential equations and finding solutions to systems of linear equations.

Can the Wronskian be used for more than two functions?

Yes, the Wronskian can be calculated for any number of functions. In general, if the Wronskian is non-zero for a set of n functions, then they are linearly independent. However, if the Wronskian is equal to zero, then the functions may or may not be linearly dependent.

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