Eliminating Arbitrary Constants

[joeljr]

Help me please eliminating im having a hard time, need full solution really appreciate it well

 

[tkhunny]

What tools are available to you? What unit are you studying? Show us something so we can have a clue how to help you.

Lacking a clue, take three derivatives on #4 and see if anything makes sense.

 

[Deleted member 4993]

Help me please eliminating im having a hard time, need full solution really appreciate it well

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Please share your work/thoughts about this assignment

The first problem has two constants - it probably is referring to a second-order linear Differential equation.

The second problem function is non linear in 'y'. What would it tell us about the original function f(x).

 

[HallsofIvy]

Think about how you solve systems of algebraic equations. If you have three equations in three unknowns, say, x, y, and z. You would add or subtract a multiple of one equation from a multiple of another to eliminate one unknown at a time. In the first problem we have two "unknowns", A and B, we want to eliminate. So we need three equations! We already have one equation so we need two more. We get those by differentiating!

In the first problem you have $\displaystyle y= Ae^x+ Be^{2x}+ \frac{1}{2}e^{-x}$. Since this has to do with differential equations, start by differentiating!
$\displaystyle y'= Ae^x+ 2Be^{2x}- \frac{1}{2}e^{-x}$
$\displaystyle y''= Ae^x+ 4Be^{2x}+ \frac{1}{2}e^{-x}$.

Now eliminate A and B from those three equations. The first thing I notice is that all three equations contain $\displaystyle Ae^x$. Subtracting the y' equation from the y'' equation eliminates A: $\displaystyle y''- y'= 2Be^x+ e^{-x}$. Subtracting the y equation from the y'' equation also eliminates A: \(\displaystyle y''- y= 3Be^{2x}.

Now we have $\displaystyle y''- y'= 2Be^x+ e^{-x}$ and $\displaystyle y''- y= 3Be^{2x}$. We can eliminate B by subtracting 3 times the first of those equations from 2 times the second. $\displaystyle 3y''- 3y'- 2y''+ 2y= y''- 3y'+ 2y= 3e^{-x}$

The second problem, $\displaystyle y^2= 4C(x+ C)= 4Cx+ 4C^2$ has only one constant so we only need one more equation. Differentiating, $\displaystyle 2yy'= 4C$. Okay, it's a little harder to see how we can (elegantly) eliminate C from those two equations so I would try "brute force". From $\displaystyle 2yy'= 4C$ we have $\displaystyle C= \frac{yy'}{2}$. Replace C in the first equation by that: $\displaystyle y^2= 2xyy'+ y^2(y')^2$.\)