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# First Order Differential Equations, Partial DE's/ Linear Dependence/Independance

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1. Find solutions to the given Cauchy- Euler equation
(a) xy'+ y =0 (b) x2y'' + xy'+y =0 ; y(1) =1, y'(1) =0

2. Find a solution to the initial value problem
x2y' + 2xy = 0; y (1) = 2

3. Find the general solution to the given problems
(a) Y' + (cot x)y = 2cosx (b) (x-5)(xy'+3y) = 2

4. Solve the Bernoulli equation y' = xy3 - 4y

5. Use separation of variables to solve the verhulst population problem

N' (t) = (a-bN) N, N (0) = N0; a,b > 0

6. Verify that each of the given functions is a solution of the given differential equation, and then use the Wronskian to determine linear dependence/ independence
Y''' - y''- 2y' = 0 {1, e-x, e2x}

https://brainmass.com/math/partial-differential-equations/first-order-differential-equations-partial-547247

## SOLUTION This solution is FREE courtesy of BrainMass!

1a.
The equation is:
(1.1)
We "guess" a solution in the form:
(1.2)
We substitute this back in the equation:

(1.3)
Thus the solution to (1.1) is:
(1.4)
Where C is a constant to be determined from initial conditions.

1b.
The equation is:
(1.5)
We "guess" a solution in the form:
(1.6)
We substitute this back in the equation:

(1.7)
We have a single pure complex root with multiplicity of two Thus the solution to (1.5) is a linear combination
(1.8)
From initial conditions:

Thus:
(1.9)

2.
The equation is:
(1.10)
Note that we can write the left hand side as:
(1.11)
Thus, the equation is:
(1.12)
Integrating both sides:
(1.13)
Where C is a constant.
The general solution is:
(1.14)
Applying initial conditions:
(1.15)
The solution is:
(1.16)

3a.
The equation is:
(1.17)
We rewrite the equation by multiplying both sides by sin(x):
(1.18)
And we note that the left hand side can be written as:
(1.19)
And the right hand side is:
(1.20)
Therefore:
(1.21)
Integrating both sides:

(1.22)

3b
The equation is:
(1.23)
We rewrite the equation:
(1.24)
The integration factor is:
(1.25)
We multiply both sides by and we get:
(1.26)
The left hand side is now:
(1.27)
We integrate both sides:

(1.28)
Where c is an integration constant.

We use partial fraction conversion:

(1.29)
Equating the coefficients of the numerator:

(1.30)
Therefore:

(1.31)

4.
The equation is:
(1.32)
Rewriting:
(1.33)
This is a Bernoulli with n=3, so we set:
(1.34)
Thus:
(1.35)
Hence:

(1.36)
This is a simple first order equation.
The integration factor is:
(1.37)
We multiply both sides by the integration factor:

(1.38)
Integration of both sides yields:
(1.39)
Integrating by parts:
(1.40)
Thus:

(1.41)
So now we can use (1.34):

So the solution to (1.32) is:
(1.42)

5.
The equation is:
(1.43)
Separating the equation:
(1.44)
Now each side is a function of a single variable, so we can integrate.
We use partial fractions:

Equating the numerators:
(1.45)
Therefore:
(1.46)

So when we integrate both sides of (1.44) we get:

(1.47)

Applying initial conditions:

(1.48)
Thus

(1.49)

6.
The equation is:
(1.50)
Solution 1:
Then indeed it is a solution since
(1.51)
Solution 2:
Then:
(1.52)
Plugging it back into the equation yields:

So it is a solution.

Solution 2:
Then:
(1.53)

Plugging it back into the equation yields:

So is a solution.
The Wronskian is:
(1.54)
Since there is no where , all the solutions are independent.

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