2ND semester Diploma students Engineering mathematics-II Important questions of new syllabus

Added: 2026-05-25

[00:00:00]Very good morning.

[00:00:01]My dear second semester degree class students in our states Today, I am uh taking your doubt clearing class.

[00:00:21]Take it easy. I am discussing one by one.

[00:00:28]So, then y is equal to a cos x plus b sin x is a solution of differential equation d squared y by dx squared plus y is equal to zero.

[00:00:37]The equation is given by y is equal to a cos x plus b sin x.

[00:00:41]Derivative of dy by dx is equal to d by dx a cos x plus b sin x.

[00:00:47]a is a constant term, a. Derivative of cos x with respect to x.

[00:00:53]b Derivative of sin x with respect to dx.

[00:00:58]You know that derivative of cos x is equal to minus sin x.

[00:01:03]Since derivative of cos x is equal to minus sin x, therefore I am writing as minus a sin x.

[00:01:10]Derivative of sin x is equal to cos x, therefore b cos x.

[00:01:14]Then again derivative d squared y by dx squared is equal to minus a constant term derivative of a sin x plus b derivative of a cos x.

[00:01:25]Derivative of sin x is equal to uh cos x.

[00:01:29]Therefore minus a into cos x plus b derivative of cos x is equal to minus sin x.

[00:01:36]Therefore this term becomes minus a cos x minus b sin x is equal to taking common minus a cos x plus b sin x. You know that a cos x plus b sin x is equal to y.

[00:01:52]Therefore I am writing as d squared y by dx squared minus y or d squared y by dx squared plus y is equal to zero. This is the >> y or d squared y by dx squared plus y is equal to zero. This is the required answer.

[00:02:07]Then e to the power cos squared x sin 2x dx.

[00:02:10]Let cos squared x t cos squared x is equal to t.

[00:02:14]Then 2 cos x into derivative of cos x is equal to dt. Therefore, 2 cos x into derivative of cos x is equal to minus sin x dx is equal to dt.

[00:02:27]Therefore, minus 2 sin x into cos x dx is equal to dt.

[00:02:32]You know that sin 2x is equal to 2 sin x into cos cos x.

[00:02:37]Therefore, this term becomes minus sin 2x dx is equal to dt.

[00:02:43]Or sin 2x dx is equal to minus dt.

[00:02:46]Therefore, integration of minus e to the power t into dt.

[00:02:51]Minus e to the power t integration of e to the power t plus c.

[00:02:55]Therefore, minus e to the power cos squared x plus c. This is the required answer.

[00:03:01]Then dx by 1 plus e to the power minus x integration.

[00:03:06]We can write as dx by 1 plus 1 by e to the power x because e to the power minus x is equal to 1 by 1 e to the power x.

[00:03:16]Then dx by taking common LCM ex ex plus 1.

[00:03:23]Therefore, e to the power x dx by 1 plus e to the power x integration.

[00:03:28]Putting 1 plus e to the power x is equal to t.

[00:03:32]Then Then tan inverse x dx.

[00:03:41]It is the integration by parts. First first function tan inverse x into derivative of integration of second function minus integration of derivative of first function integration tan inverse x minus 1 plus x squared dx again x tan inverse x equal to t 2x dx is equal to dt by 2. Therefore, I'm writing as dt.

[00:04:06]x tan inverse x minus 1 by then integration of dt by t is equal to log t.

[00:04:14]Therefore, x tan inverse x minus 1 by 2 log 1 plus x squared plus c this is the required answer.

[00:04:21]Then this equation is this problem is very important. 0 to pi by 2 sine root over sine x by root over sine x plus root over cos x dx Let I is equal to 0 to pi by 2 sine x root over by sine x root over plus cos x root over dx 1 then putting the value pi by 2 minus x pi by 2 minus x pi by 2 minus x. You know that sine 90 minus x is equal to cos x.

[00:04:52]sine pi by 2 means sine 90 minus x root over cos sine x dx this is the equation.

[00:04:58]I don't know 1 or 2 we have pi by 2 root over sine x by root over sine x plus root over cos x dx plus root over cos x by root over cos x plus root over sine x dx Then adding we have root over sine x plus root over cos x is equal to root over sine x plus root over cos x dx This term becomes 1 and this term cancel.

[00:05:20]Therefore, 0 to pi by 2 dx You know that integration of dx is equal to x.

[00:05:27]What do you mean by pi by 2? That means pi by 2 minus 0.

[00:05:31]Or 2i is equal to pi by 2 or i is equal to pi by 2 by half is equal to pi by 4 this is the required answer.

[00:05:39]Then dy by dx is equal to y plus 2 or dy by dx is equal to y plus 2 dy is equal to y plus two into dx.

[00:05:49]y by [snorts] two into dx or dy by y plus two is equal to dx.

[00:05:54]Integrating both sides we have dy by y plus two is equal to integration of a dx.

[00:06:00]Let y plus two is equal to t.

[00:06:03]Then derivative y plus two is equal to dy is equal to dt. I mean dy is equal to integration of a dx.

[00:06:11]Then dt by t x log y plus two is equal to x plus c.

[00:06:17]Or log y plus two is equal to x plus c.

[00:06:20]Or we can write y plus two is equal to e to the power x plus c.

[00:06:24]Or y is equal to e to the power x plus c minus two is the required answer.

[00:06:30]Then one plus cos two x dx. One number of a variable is cos squared x minus sin squared x. dx root over. Then sin squared x minus sin squared sin squared x cancel.

[00:06:42]Or cos squared x plus cos squared x is equal to two cos squared x.

[00:06:48]You know that root over Uh you know that two cos squared x and a root over root two cos x dx. Then integration of root two cos x dx. root two cos x Since root integration of a cos x is equal to sin x is the required answer.

[00:07:08]This problem is very important.

[00:07:10]Uh root over integration of a e to the power x sin x dx.

[00:07:14]Let I is equal to e to the power x sin x dx.

[00:07:18]First first function sin x into integration of a second function.

[00:07:22]minus integration of derivative of derivative of first function into integration of a second function e to the power x dx dx.

[00:07:33]sin x into e to the power x e e to the power x dx integration of a e to the power x minus derivative of a sin x is equal to cos x.

[00:07:42]Integration of a ex is equal to Then e to the power of x sine x minus integration by double integration by cos x into integral of our second function e to the x dx minus derivative of our cos first function cos x by dx into integration of our e to the x dx into dx e to the x into sine x minus cos x integration of our e to the x e x minus derivative of our cos x minus sine x into e to the x dx that e to the x into sine x minus cos x into e to the x plus sine integration of our sine x into e to the x dx or we can write that e to the x into minus minus e to the x cos x minus integration of our sine x into e to the x dx and I is equal to you know e to the x x into sine therefore I'm writing like that I or I plus I 2 I is equal to e to the taking common e to the x sine x minus cos x or 2 I is equal to e to the x within bracket sine x minus cos x or I is equal to e to the x by 2 within bracket sine x minus cos x plus c is the required answer and 3 x plus 1 by x plus 1 into x minus 2 dx we have 3 x plus 1 by x plus 1 into x minus 2 is equal to a by x plus x x minus 2 then 3 3 x plus 1 by x plus 1 into x minus 2 is equal to uh taking LCM a into x minus 2 plus b into x plus 1 this term become cancelled 3 x plus 1 is equal to a within bracket x minus 2 plus b with x This term becomes 3 into 2 + 1.

[00:09:43]Is equal to A into 2 minus 2 plus B into T T 2 plus 1.

[00:09:49]Or 6 plus 1 is equal to 0 plus 3 B.

[00:09:55]Or 7 is equal to 3 B or B is equal to 7 by 3.

[00:10:01]term becomes zero.

[00:10:05]Therefore, 3 into minus 1 plus 1 is equal to A This term becomes zero.

[00:10:12]Or >> Mhm.

[00:15:24]>> Mhm.

[00:17:19]>> Mhm.

[00:19:59]>> Mhm.