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# Prove a substitution converts a given second order equation to a first order equation

1. Consider the second-order differential equation

and let be a solution to the equation. Show that the substitution converts the equation

into a first-order liner equation for .

2. By inspection, find a nonzero solution of the second order differential equation

and use part (a) to find a solution of the differential equation

with

Incomplete.

1. Evangelos says:

Every solution comment I post disappears. WordPress go to hell.

2. Evangelos says:

a.) If we make the substitution , with being a nonzero function satisfying

The equation

Becomes

Grouping terms by multiples of v and its derivatives gives us

Since u satisfies , we get

And since is a nonzero function, we can divide both sides by it, giving us

If we let and , the above equation becomes

Which is a linear first-order equation of v’, as requested.

3. Evangelos says:

a.) If we make the substitution , with being a nonzero function satisfying

The equation

Becomes

Grouping terms by multiples of v and its derivatives gives us

Since u satisfies , we get

And since is a nonzero function, we can divide both sides by it, giving us

If we let and , the above equation becomes

Which is a linear first-order equation of v’, as requested.

b.) By inspection, is a nonzero function that satisfies

Now, if we use the method of substitution from part (a.), the equation

Becomes

And since we know that , the above equation becomes

Which is a linear equation of . If we let

Then

Now, we can deduce the value of v'(0) using our given information. We know that , and we also know that when , , and , so must be 0. Plugging in those values gives us when . Thus:

But if that is the case, then we can integrate to give us

And since , we can multiply to get

4. Evangelos says:

Testing inline latex formatting should be x = 0

• Evangelos says:

I typed out an entire reply and it disappeared D:

• Evangelos says:

Testing both inline and separate line

I’m still salty my post got lost

5. Evangelos says:

And the equation

Becomes

Now, if we group together terms by v and its derivatives, we get

But, we know from our given information that u satisfies

So the factor multiplying v becomes 0, giving us

Since u is a nonzero function, we can divide both sides by it, giving us

Now, if we define two new functions of x as follows

The above differential equation becomes

Which is a first-order linear equation of v’

b.) By inspection, we are to find a nonzero function u = f(x) that satisfies the equation

The following function satisfies the equation

Now, we are to use the function u and the information from part (a.) to find a solution to the following equation

From part (a.), we know that we can use the substitution to turn the above equation into a first-order linear equation of , as follows

And of course, from our inspection, we know that , so

And we know that , so the above equation becomes

And since the exponential function is never 0 for any real x, we can divide both sides by to give

Now, since this is a linear equation of v’, we can use the equation from Theorem 8.3, along with our given values when , to give an equation for v’. Let

Our equation for becomes

Going back to our given information, we know that or , so when , since and . As such, . Giving us

Integrating this gives us

And since we know that , the constant C must be 1. Now, to get y, we go back to our original substitution