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### Course: Differential equations>Unit 2

Lesson 1: Linear homogeneous equations

# 2nd order linear homogeneous differential equations 2

Let's find the general solution! Created by Sal Khan.

## Want to join the conversation?

• At , Sal suggests the function e^(rx) to satisfy the equation since it keeps the derivatives "in line", so to speak. That all makes sense to me, but why not generalize it even further to the set of functions e^(rx+k), where k is some arbitrary constant? Wouldn't that be equally valid while covering even more solutions?

Consequentially the general solution to the diff equation would be y(x) = C_1e^(r_1x + k_1) + C_2e^(r_2x + k_2). Wouldn't that work equally well while covering more answers?
• Nevermind, I realized that the more complicated version doesn't add any solutions since e^(rx + k) = e^(rx) * e^k. The e^k-factor is constant, so the term C_1*e^(rx) already covers the span of C_2*e^(rx + k) - you can just pick the constant C_1 to equal C_2*e^k, and the terms are equal!
• So how do you solve for r if instead of 5 and 6? You just have p and q? y"+py'+qy=0.
• I'm not sure what you mean, but I would guess that you would have to solve for r but putting p and q into the quadratic formula. It might get messy, though.
• Are the solutions to ALL 2nd order homogenous equation y=e^x functions?
• If r is imaginary, they will actually be a sum of trigonometric functions, not exponential functions, because e^(ix) = cos(x) + i*sin(x).
• How can we proof these are the only solutions? meaning where I can find a method or way to decide these are the only solutions to these problem.
• at , why sal multiplied c's to the answers, would it not bean answer withoat c's ?
• It would have been A solution but he's getting the general solution.
I believe the previous video has the properties he used.
• If instead of that equation equaling 0 it equaled a constant such as 18, would the function still be homogeneous. as it could still be written as y''+5y'+6y-18=0 or is my reasoning incorrect.
• Equation y''+5y'+6y=18 is not homogenous. I believe it can be sold by method of undetermined coefficients (presented further in differential equations course). Shortly, the result of equation should be threated like 18+0, so the general solution would be general solution to this equation =0 plus the particular solution to the same equation =18
• How do we know there aren't other solutions, apart from the exponential one? I trust Salman Khan, but I would like to know the proof. :)
• for 1st order, like y'=ky, the proof is quite easy: let y(t) be any solution (perhaps not the exponential one?), then consider the function u(t) = e^(-kt) y(t)

check that u'(t) = -k e^(-kt) y(t) + e^(-kt) y'(t) by product rule, now since y‘(t) = k y(t), so u'(t) = (-k + k) e^(-kt) = 0!, what can u(t) be? u(t) = C, a constant, so the unknown solution y(t) has the property that e^(-kt) y(t) = C, so y(t) = C e^(-kt) is certain! the proof based on the fact that u'(t) = 0 implies u(t)=C, this is a basic calculus fact, that can further be proved using even more elementary real analysis techniques, however, if you accept this, then the proof for first order linear equations goes like what I've shown.
• Sal is giving the solution of second order differential equation through the method of finding complimentry function (C.F.) or complete solution.
But can anyone tell me how we get to the result of exponential function?
• Given a standard linear differential equation:
y' + p(x)y = q(x) your integrating factor will always be e^(integral of p(x) dx). You then multiply through by this.
• I don't understand how Sal applied the chain rule to this function. Can someone please explain it to me step-by-step?
• The only time he used the chain rule was after substituting `y = e^(rx)`. The chain rule states that given:
`y = f(g(x))`
`y′ = g′(x) * f′(g(x))`
Substituting the equation in the problem:
`y = e^(rx)`
`f(x) = e^x`
`g(x) = rx`
`f′(x) = e^x`
`g′(x) = r`
`y′ = r*e^(rx)`
In the video Sal then takes the second derivative using the same process:
`h(x) = r*e^x`
`h′(x) = r*e^x`
`k(x) = rx`
`k′(x) = r`
`y′ = h(k(x))`
`y′′ = k′(x) * h′(k(x))`
`y′′ = r*r*e^(rx)`
`y′′ = r²e^(rx)`