Enable contrast version

# Tutor profile: Chinmay M.

Inactive
Chinmay M.
Tutor for four years in subjects like mathematics and chemical engineering
Tutor Satisfaction Guarantee

## Questions

### Subject:Calculus

TutorMe
Question:

Show that y= A sin 2x + 3 cos 2x is solution of y" + 4y= 0

Inactive
Chinmay M.

We have a second order differential equation and we have been given the general solution. Our job is to show that the solution is correct.We will do this substituting the answer into the original 2nd order differential equation. We need to find the second derivative of y y= A sin 2x + 3 cos 2x ..........................(1) First Derivative= y' = 2* (A cos2x) - 2 * ( 3sin2x)= 2A cos2x - 6 sin2x Second derivative= y"= - 2* (2A sin2x) - 2*(6cos2x)= -4A sin2x - 12cos2x ...........(2) Now substitute back in y"+ 4y. To be the solution it should come to zero. LHS= y"+ 4y= -4A sin2x - 12cos2x + 4*(A sin 2x + 3 cos 2x) ...............From (1) & (2) =-4A sin2x - 12cos2x + 4A sin2x+12cos2x = 0 Hence y= A sin 2x + 3 cos 2x is solution of y" + 4y= 0

### Subject:Calculus

TutorMe
Question:

Differentiate ln ( sinx -cotx)

Inactive
Chinmay M.

Let f(x)=ln ( sinx -cotx) then derivative of f(x) will be, f'(x) = (1/sinx-cotx) *(cosx + csc^2 x) .... (1) For differentiating logarithm , differentiating the logarithm (i.e. the outside function) you need to substitute the inside function into the derivative and then multiply by the derivative of inside function Here sinx - cotx is inside function so the answer is (1)

### Subject:Algebra

TutorMe
Question:

Find the quadratic equation with real coefficients which has 3 - 2i as a root (i = √-1).

Inactive
Chinmay M.

According to the problem, coefficients of the required quadratic equation are real and its one root is 3 - 2i. Hence, the other root of the required equation is 3 - 2i (Since, the complex roots always occur in pairs, so other root is 3 + 2i. Now, the sum of the roots of the required equation = 3 - 2i + 3 + 2i = 6 And, product of the roots = (3 + 2i)(3 - 2i) = 322 - (2i)22 = 9 - 4i22 = 9 -4(-1) = 9 + 4 = 13 Hence, the equation is x22 - (Sum of the roots)x + product of the roots = 0 i.e., x22 - 6x + 13 = 0 Therefore, the required equation is x22 - 6x + 13 = 0.

## Contact tutor

Send a message explaining your
needs and Chinmay will reply soon.
Contact Chinmay

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage