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

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.

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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:

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Thus:

(1.9)

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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)

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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)

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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.

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We use partial fraction conversion:

(1.29)

Equating the coefficients of the numerator:

(1.30)

Therefore:

(1.31)

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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):

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So the solution to (1.32) is:

(1.42)

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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)

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Applying initial conditions:

(1.48)

Thus

(1.49)

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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.

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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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