solve differential equation with substitution

Comments · 68

• CH

  @chuaprincecarl9845 · 5 years ago

  the marker switch is smooth af, Michael Jackson is proud.

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• DE

  @delteide1379 · 10 months ago

  8 years later and you're video is still helpful, thy

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• SA

  @sam-kx3ty · 6 years ago

  You’re one of the best math lecturers in the world please keep it up .

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• RA

  @RaceAndRetreat · 6 months ago

  Video is 9 years old and still helping people. Such a nice concise explanation. Keep up the amazing work.

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• CA

  @candlelightc4699 · 3 years ago

  its 5 years later but thank you for the very clear explanation

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• IV

  @ivypellerin3166 · 5 years ago

  Thank you for showing how we get the substitution for dy/dx my profs like to skip intermediate steps also loved the flawless marker flipping hahaha

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• 79

  @79Shotinthedark · 8 years ago

  This is my first time watching one of your videos. I appreciate how you take your time with the problem and that you write very clearly (surprisingly hard to find). Using the two colors made it easier to follow. I learn a lot of my math from Youtube and this was very helpful. Thank you.

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• HO

  @HonsHon · 6 years ago

  Thank you! Helping me so much in preparing for the final in my DE class. Ever since I was in Calc 2 I have been watching these, and they are so helpful.

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• AG

  @ageofkz · 9 years ago

  Is there a special name for this sort of functions where you make a substitution to solve it? <br>For example, homogenous 1st ODE you will substitute f(y/x)=f(v), v=y/x.

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• CH

  @ChefSalad · 7 years ago

  Without WolframAlpha, I know how to solve for v. Start by taking the exponential function of both sides, and relabeling the c: sec(v)+tan(v)=C₁*e^(-1/x). Change sec and tan to sin and cos and combine fractions: (1+sin(v))/cos(v)=C₁e^(1/x). Shift the sin and cos to cos and sin: (1+cos(v+π/2))/sin(v+π/2)=C₁e^(−1/x). Reciprocate: sin(v+π/2)/(1+cos(v+π/2))=C₂e^(1/x). Use the tangent half-angle identity: tan(v/2+π/4)=C₂e^(1/x). Take the arctan of both sides. v/2+π/4=arctan(C₂e^(1/x)). Solve for v: v=2arctan(C₂e^(1/x))−π/2. Substitute back in v=y/x²: y/x²=2arctan(C₂e^(1/x))−π/2. Solve for y: y=2x²arctan(C₂e^(1/x))−πx²/2. BAM! Solved for y.

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• DA

  @danillatency · 1 year ago

  Wow, it&apos;s very clear, thank you!

• DO

  @DougCube · 9 years ago

  Here is the closed-form solution: 2(x^2)arctan(tanh((Cx-1)/(2x))). Not that anyone cares...

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