I get your point, 1/4 * e^x(c1) , you can express c2 as 1/4 * c1, so that you have e^x(c2). Because it's a constant, it does not really change anything, in short you can as well express it as how you did, it's also correct. Nice observation, thumbs up
This is exhaustive. I never saw this in the lecture hall
I see
Well understood
x^2/(xy-x^2) kindly solve this homogeneous equation.
Sir. How can you solve a question like these using your formulae.
Y'-³/x Y= Y⁵
Have you tried your hands on it already?
Amazing explanations
Thanks so much Emmanuel
boss abeg the Q2 the part wey you substitute the expressions and n why say the 1/2 from the integration part dey affect the constant c too
Can you please come again with your question, and kindly state the time in the video, for easy reference. Thank you
@@SkanCityAcademy_SirJohn at 16:50 I don't think the 1/2 should affect the constant too
The last part at the constant, there is a little confusion time of the video=27:25. The answer supposed to be y= ((sinx-cosx)/4 +ce^x )^2
I get your point, 1/4 * e^x(c1) , you can express c2 as 1/4 * c1, so that you have e^x(c2). Because it's a constant, it does not really change anything, in short you can as well express it as how you did, it's also correct. Nice observation, thumbs up
Thanks for the clarification. You're the best 👑
@@iindongotaapopi1218 what's your nationality?
@@SkanCityAcademy_SirJohn 👑
@@SkanCityAcademy_SirJohn Namibian
Thank you
You are most welcome. Keep watching
❤
Thanks so much Cossa