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

• roymkim
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.

#### roymkim

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

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

f g'-f' g=t2(g/t)'=t2et

I'm still not getting the right answer.

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

Where are you getting that t-1? It clearly does not belong.

, 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.