Math 24 Section 2.7: Substitution

Added: 2026-05-15

[00:00:01]section 2.7 substitutions uh so far we learned uh several ways for solving differential equations and in this section we can to look at at some specific uh differential equations that didn't work no matter which method you use that we have learned so far um so so far we learned separable differential equations where you separate the variables homogeneous differential equations exact linear Peri beri so these are the ones that we learned so far but sometimes you come across a differential equation that none of these work so what we do is there's a specific or special way of handling these kind of uh equations and let me show you by starting with an example if we have this differential equation we asked to solve it so to solve such a different differential equation the first thing you can think about is [Music] um if Ln the natural log is not there what happens if natural log is not there then it's possible to be sold using what we learned so far but since Len is there it's not it's um it makes the differential equation more difficult but look at this trick how this trick will make this equation easier you can let you equal that extra or additional factor which is Ln of Y and this is I'm substituting U equal natural log of Y then du DX = 1 / y * Dy DX Dy DX so from that equation that's dydx + y y natural log of y = y e to thex so what I'm going to do from this equation here I'm going to solve for Dy DX which is y du DX so everything is going to be transformed from Dy DX into d u DX this would give us y du DX plus y this is U equals keep this the same now just looking at this one thing you can see that we can do is divide everything by y you get D DX plus u = e to X so I was able to transform everything from uh YX into ux now the new form looks easier right and to be able to solve this equation we use the integrating Factor e raed to P of X DX which is one here e to DX and we can multiply the entire equation x e to the x d DX e x * e x is e to the 2x okay so now uh this equation can be Rewritten as DDX of e to x * u = e to 2X then we integrate both sides this one the integral drops the derivative this is e to the 2X and 12 if you use the U sub U = 2x you get 12 e to the 2x plus a constant C divide everything by e to the X this becomes e to the x c e to thex but then remember U is Ln of Y or natural log of Y 12 e to the X plus c e to thex you can leave it like this or you can do e of that to find the Y I would I usually just leave it like that any questions about this we'll go for one more example but I want you to try it first on your own x² e tox so solve this differential equation so to solve this one um one idea would be to substitute to equals cosine y now D DX is netive sin y Dy DX so then the given equation becomes X see instead of s y d Dy DX we put a negative du so negative du DX is negative sin y Dy DX so it's negative D DX uh plus u = x² X notice everything is a as a function in U and X now divide everything by negative X we get du DX - 1 /x U is x e to the x now we can use the integrating Factor mu = e the integral of -1 /x DX e natural log of x e natural log of x to the1 X to1 or 1 /x that's the integrating Factor now we can multiply everything by 1 /x which is e x then we have DDX of 1 /x * u = e to the X integrating both sides a x + C now multiply everything by X so x e x plus c x but remember U is cosine y so cine y = x e to x + c x you can leave it like that anyone has any questions about this no questions