Differential Equations

Get insights from 136 questions on Differential Equations, answered by students, alumni, and experts. You may also ask and answer any question you like about Differential Equations

Follow Ask Question
136

Questions

0

Discussions

3

Active Users

0

Followers

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

line: 3y – 2z – 1 = 0, 3x – z + 4 = 0

a p o i n t P ( t 4 3 , 2 t + 1 3 , t )  on the line, Q (2, -1, 6)

P Q 2 = 2 9 ( 7 t 2 5 6 t + 2 2 0 )

( P Q ) m i n = 2 6  

 

New answer posted

2 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

dydx=yx (4y2+2x2) (3y2+x2)

Put y = vx

dydx=v+xdvdx

v+xdvdx=v (4v2+2) (3v2+1)

ln (y28+y2)=2ln2y38+y2=4

[y (2)]=2

n=3

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

y 2 d x + ( x 2 ? x y + y 2 ) d y = 0

d x d y + x 2 ? x y + y 2 y 2 = 0 ? d x d y + ( x y ) 2 ? ( x y ) + 1 = 0

Put x = vy ? v + y d v d y + v 2 ? v + 1 = 0

? y d v d y + v 2 + 1 = 0

? ? 6 + l n | y | = ? 4

l n | y | = ? 1 2

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

y 2 d x + ( x 2 ? x y + y 2 ) d y = 0

d x d y + x 2 ? x y + y 2 y 2 = 0 ? d x d y + ( x y ) 2 ? ( x y ) + 1 = 0

Put x = vy ? v + y d v d y + v 2 ? v + 1 = 0

? y d v d y + v 2 + 1 = 0

? ? 6 + l n | y | = ? 4

l n | y | = ? 1 2

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Letx3=θθ2 (π4, 3π4)

y=tan1 (secθtanθ)

tan1 (1sinθcosθ)

dydx=3x22d2ydx2=3x

x2d2ydx26y+3π2=0x2y116y+3π2=0

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  f ( r 4 ) = 2 , f ( r 2 ) = 0 & f ' ( r 2 ) = 1

g(x)   ( f ' ( t ) s e c t + t a n t s e c t f ( t ) ) d t

= [ f ( t ) s e c t ] x π / 4

=   2 * 2 f ( x ) s e c x

= 2 c o s x f ( x ) c o s x

=   2 s i n x + 1 s i n x = 3

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

 dydx+2xx1y=1 (x1)2

IF =e2xx1dx

e2x (x1)2

ye2x (x1)2= {e2x (x1)2 (x1)2dx+C

y = e2x2 (x1)2+C (x1)2

y (2) = 1+e42e4, C=12

y (3) = eα+1βeα=e6+18e6

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

dy1+y2=2exdx1+ (ex)2+C

tan1y=2tan1ex+C

x = 0, y = 0

0=2tan1+C

C=+π2

now at x = ln3

tan1y=2tan1 (eln3)+π2

6 (y' (0)+ (y (ln3))2)=6 (1+13)=4

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Let p1 : y2 = 8x

p2 : y2 = 16 (3 – x) = -16 (x – 3)

finding their intersection points.

y2 = 8x & y2 = -16 (x – 3)

8x = -16x + 48

= 2 . 0 4 ( 3 y 2 1 6 y 2 8 ) d y

= 2 ( 3 y y 3 3 * 1 6 y 3 3 * 8 ) 0 4 = 16

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

GOF is differentiable at x = 0

So R.H.D = L.H.D.

d d x ( 4 e x + k 2 ) = d d x ( ( | x + 3 | ) 2 k 1 | x + 3 | )                

⇒ 4 = 6 – k1 Þ k1 = 2

Now g (f (-4) + g (f (4)

= 2 (2e4 – 1)

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.