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First Order Differential Equations and Values of Parameters

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1. Given satisfying , ; &#960; ; find

2. Given such that , ; list all possible solutions. For which of these does ?; ?

3. Suppose for it is known that

Where "a" is a parameter. Determine the value of this parameter which ensure the existence of a relation such that , C a constant, for all satisfying (1) and then deduce . What is C if .

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1. Given satisfying , ; π ; find
This is linear equation. When compared to the general form of the linear equation , where P and Q are functions of x, we have . The solution for such equations is given by where C is arbitrary constant.
Plugging the values of P and Q in the above equation, we get the solution as
Hence the solution is given by . Given that π ; Plugging the value x=π in the above solution, we see that
. So the solution for the given equation is
2. Given such that , ; list all possible solutions. For which of these does ?; ?
Rewriting the given equation as and dividing by y2 throughout, we get . Let . This gives . Hence the equation transforms as . This is a linear equation. Writing the solution as in the above problem, we have

Plugging the value , we get the solution to the original equation as .
For y(e)=0, we can not find any value for C, whereas for y(e)=e, we see that 1=-lne+C which implies, C=2

3. Suppose for it is known that

Where "a" is a parameter. Determine the value of this parameter which ensure the existence of a relation such that , C a constant, for all satisfying (1) and then deduce . What is C if .

Before proceeding for the solution, we recollect the definition of an exact differential equation and method to write its solution.
A differential equation of the form Mdx+Ndy=0 where M and N are functions of x and y is called an exact equation, if , where represents the partial derivative of M with respect to y treating x as constant. Then the solution for the differential equation is given by where the first integral is evaluated with respect to x treating y as constant and in the second integral only those terms which are not containing x are integrated with respect to y.

So, considering the given equation to have relation , it needs to be solved. This can happen if the equation is exact. So, we have .
For this to be exact, we need to have
Carrying out the differentiation, we see that . For this to be true for all x and y, we need to have a=2.
Hence we have the equation
Hence the solution is given by
Carrying the integration as explained above, we get

Hence in the above required solution.
If y(0)=1, from we have

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