Homogeneous Differential Equations

This calculus video tutorial provides a basic introduction into solving first order homogeneous differential equations by putting it in the form M(x,y)dx + N(x,y)dy = 0. You can solve it by substitution using the equation y = vx and dy = xdv + vdx. Following the change of variables, you can integrate the differential equation by separation of variables. This video contains plenty of examples and practice problems of finding the general solution of homogeneous differential equations including solving the initial value problem.
Improper Integrals:
• Improper Integrals - C...
Integration Into Inverse Trig:
• Integration into Inver...
Integration of Rational Functions:
• Integration of Rationa...
Integral of Logarithmic Functions:
• Integral of Logarithmi...
Integrating Exponential Functions:
• Integrating Exponentia...
Integration Formulas:
• Integration Formulas F...
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Integration By Tables:
• Integration By Tables
Reduction Formulas - Integration:
• Reduction Formulas For...
Center of Mass & Moments:
• Moment, Center of Mass...
Center of Mass & Centroid Problems:
• Center of Mass & Centr...
Hydrostatic Force Problems:
• Hydrostatic Force Prob...
Probability Density Functions:
• Probability Density Fu...
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Normal Distributions - Calculus:
• Normal Distribution & ...
Homogeneous Differential Equations:
• Homogeneous Differenti...
1st Order Linear Differential Equations:
• First Order Linear Dif...
Bernoulli's Equation for Differential Equations:
• Bernoulli's Equation F...
Slope Fields:
• Slope Fields | Calculus
___________________________________
Converging & Diverging Sequences:
• Converging and Divergi...
Monotonic & Bounded Sequences:
• Monotonic Sequences an...
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[@TheOrganicChemistryTutor]

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Professor Organic Chemistry Tutor, thank you for a powerful analysis on First Order Homogeneous Differential Equations. This is where Techniques of Integration in Calculus Two is used for finding solutions to Ordinary Differential Equations. Differential Equations is an extension of Calculus. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

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[@Mathefusions18]

The equations must have same dimensions for them to be homogeneous. Great work. Thanks.

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Great work but you can also write e^c as k, because e^ to a constant maybe a constant, but it's not the same constant "c". It's different so you have to call it something different.

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Is there a video on exact differential equations?

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THANK YOU. I HOPE I CAN ANSWER THE QUIZ ON HOMOGENEOUS.

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Can you explain why we stop at abs(x) = c(x-y)^2? Why is that the desired form? Does it just happen to coincide with applications in the sciences and therefore this is useful for university students?

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[@Zwalter]

I am really confuse where should you place the "c" because it does matter on where you would place the c because you will get a completely difference answer. Lets take for example on you last example. The general solution I got was: lnx^2 = e^(y/x) +c Now when I computed the value of c, I got 0 = 1 +c c = -1 Instead of e^(y/x) = lnx^2 +c 1 = 0 + c c = 1 Please do clarify on this one because I am having confusions. Thanks!

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What did you find x in the first problem and y in the second problem?

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can someone please explain the first example isn't that a linear first order equation , so can we use the integration factor method?

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[@winner1338]

0:44 How do you come up with this equation? y = vx Why not, y = x + 1 or anything else?

[@eashanmathur2030]

Thats just the special v-subsitution you need to memorize for homogenous equations. You want to see if the function can be a function of y over x. v=y/x is the same as y=vx

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@@eashanmathur2030 ok, thank you

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Just think of "v" as a variable that allows x to always have the same value as y for all values of y.

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Win Ner the goal is to substitute v into the equation so you can turn it into a separable equation in terms of x and v and solve it like that. Then you just substitute your v=y/x back into the equation for the answer

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the second example was a second order DE

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Question I did it inversely and I got to the point where you got the anti derivative of. 1/x dx = 1/2•e^v. But at the end I got a different answer because I added the arbitrary constant C to the exponential instead of the natural log what is the explanation behind this? Did I just mess up somewhere?

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folks need to be careful at 5:34. The -1 is to be considered inside the LN(), e.g., ------> ln [ ( 1 - v )^(-1) ]

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Question, at 2:14 when dividing both sides by x, why is 2x(xdv+vdx) simplified to 2(xdv+vdx) instead of x(xdv+vdx)?

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Left side is constant, he only took out the x from the right side as a co_ factor and replace it with 1

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