Differential Equations y" + y = tanx, solve this DE

[twoski]

Homework Statement

1. y" + y = tanx, solve this DE.

2. dP/dt = P(1 - P) where P = (c₁e^t / 1 + c₁e^t), verify that P is a solution to this DE.

3. Given the pair of functionsx and y, show they solve this system:
dx/dt = x + 3y
dy/dt = 5x+3y

x = e^-2t + 3e^6t
y = -e^-2t + 5e^6t

The Attempt at a Solution

For 1, we want to get x and y on the left and right hand sides of the equation so we can integrate i think.

so we go

dy/dx + y = tanx
dy + ydy = tanxdx
∫dy + ∫ydy = ∫tanxdx (is this right?)

For 2, the equation is not quite in standard form (dy/dx + P(x)y = f(x)), but it doesn't look like we can achieve this form. How do you proceed?

For 3, I'm not even sure how to solve it. There aren't any examples in this format...

 

[Mark44]

Homework Statement

1. y" + y = tanx, solve this DE.

2. dP/dt = P(1 - P) where P = (c₁e^t / 1 + c₁e^t), verify that P is a solution to this DE.

3. Given the pair of functionsx and y, show they solve this system:
dx/dt = x + 3y
dy/dt = 5x+3y

x = e^-2t + 3e^6t
y = -e^-2t + 5e^6t

The Attempt at a Solution

For 1, we want to get x and y on the left and right hand sides of the equation so we can integrate i think.

so we go

dy/dx + y = tanx
dy + ydy = tanxdx

No, this is wrong. Your "ydy" term on the left should be "ydx" since you multiplied both sides by dx. In any case this isn't the right approach. What you have is a first-order nonhomogeneous equation. Your general solution will consist of one part that is a solution to the homogenous equation and another part that is a particular solution to the nonhomogeneous equation. Look in your textbook to see if there are any similar worked examples.

∫dy + ∫ydy = ∫tanxdx (is this right?)

For 2, the equation is not quite in standard form (dy/dx + P(x)y = f(x)), but it doesn't look like we can achieve this form. How do you proceed?

To proceed, just show that P as defined above is a solution to the equation. You are not asked to solve the equation.

If you were, the equation is separable, meaning that it isn't too hard to get P and dP on one side and t and dt on the other. Again, though, you are NOT asked to solve the equation.

For 3, I'm not even sure how to solve it. There aren't any examples in this format...

You are GIVEN the solution, so there is nothing to solve. Just show that x(t) and y(t) satisfy the system of differential equations.

My advice to you is to pay closer attention to what is being asked for in a problem. Whenever a problem asks you to verify that such-and-such is a solution of some differential equation, you don't have to solve the equation. All you need to do is to verify that substituting the solutions function(s) into the differential equation(s) results in a statement that is identically true; i.e., for all values of the independent variable.

 

[SammyS]

Homework Statement

1. y" + y = tanx, solve this DE.

The Attempt at a Solution

For 1, we want to get x and y on the left and right hand sides of the equation so we can integrate i think.

so we go

dy/dx + y = tanx
dy + ydy = tanxdx
∫dy + ∫ydy = ∫tanxdx (is this right?)

Isn't that the second derivative of y in the given equation ?

$\displaystyle\ \frac{d^2y}{dx^2}+y=\tan(x) \$

 

[Mark44]

Good eye, Sammy. I missed the double prime (y'') in the equation.

 

[twoski]

Okay so the first question is slightly different as well.

The solution y = -(cos x) ln(sec x tan x) is given, all that is needed is to verify it is the explicit solution.

To do this, could i derive y twice then add y" to y and show that the result is tanx?

 

[Mark44]

Okay so the first question is slightly different as well.

The solution y = -(cos x) ln(sec x tan x) is given, all that is needed is to verify it is the explicit solution.

"Explicit solution" - what is that? If they have given you a solution, then that solution would be explicit, since it was "explained" to you, which is what "explicit" literally means.

To do this, could i derive y twice then add y" to y and show that the result is tanx?

Yes, in a way. If you differentiate y twice (the process to get the first and second derivative - we don't say "derive" in this context, in English), and substitute for y'' and y, the equation y'' + y = tan(x) should be identically true. That's what it means for a particular function to be a solution of a differential equation.

 

[twoski]

From my text, here is what they mean by explicit

You should be familiar with the terms explicit
and implicit functions from your study of calculus. A solution in which the dependent variable
is expressed solely in terms of the independent variable and constants is said to be an
explicit solution. For our purposes, let us think of an explicit solution as an explicit formula
y = Φ(x) that we can manipulate, evaluate, and differentiate using the standard rules

 

[Mark44]

From my text, here is what they mean by explicit

You should be familiar with the terms explicit
and implicit functions from your study of calculus. A solution in which the dependent variable
is expressed solely in terms of the independent variable and constants is said to be an
explicit solution. For our purposes, let us think of an explicit solution as an explicit formula
y = Φ(x) that we can manipulate, evaluate, and differentiate using the standard rules

Gotcha - explicit function. On the other hand, a solution given as, say, x² + y³ = 3 would represent a solution given implicitly.

 

[twoski]

Okay so i solved the first 2, all that you got to do is compute some derivatives, not too bad. The third is a bit trickier though. Do i solve them one at a time, or is there some trick i need to do?

 

[Mark44]

Okay so i solved the first 2, all that you got to do is compute some derivatives, not too bad. The third is a bit trickier though. Do i solve them one at a time, or is there some trick i need to do?

For #3, there's nothing to solve -- they have given you the solutions!

3. Given the pair of functionsx and y, show they solve this system:
dx/dt = x + 3y
dy/dt = 5x+3y

x = e^-2t + 3e^6t
y = -e^-2t + 5e^6t

For the last two equations, find, respectively, dx/dt and dy/dt. Show that dx/dt is identically equal to x + 3y, and that dy/dt is identically equal to 5x + 3y.