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```                   Find the general solution of the differential equation:( (xlogx) dy/dx ) + y = (2/x) logx
```

3 years ago

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### Answers : (1)

```                                        This is solved by integrating factor method.$\frac{\mathrm{d} y}{\mathrm{d} x}+y/(xlog(x))=2/x^2 \\=>\int 1/xlog(x)dx=log(log(x)) \\=>e^{log(log(x))}(\frac{\mathrm{d} y}{\mathrm{d} x}+y/(xlog(x))=2/x^2) \\=>e^{log(log(x))}*y=\int e^{log(log(x))}/x^2dx=\int log(x)/x^2dx=-e^{-log(x)}+c$Should be enough.Arun KumarIIT DelhiAskiitians Faculty
```
one year ago

### Post Your Answer

If P= x 3 - 1/ x 3 and Q=x-1/x , x belongs from 0 to infinity , find the minimum value of P/Q 2

P = x 3 - 1/x 3 =(x-1/x)(x 2 +1/x 2 +1) Q=x – 1/x P/Q 2 when simplified it comes in the form of (x-1/x)+3/(x-1/x) then we can apply A.M greater than or equal to G.M we get the minimum value ...

 udit maniyar 5 months ago

did u get this after simplification p/q 2 = (x – 1/x) + 3/(x – 1/x)

 udit maniyar 4 months ago

if you feel the answer is right then approve it

 udit maniyar 5 months ago
how to calculate period of sin^4x+cos^4x

the expression could be written as 1-2cos^2(x).sin^2(x) so here it is repeating afer pie thus the interval is pie approve if useful

 grenade 4 months ago

when complementry angles r there ans is period by general method /2.........when power of each is 1

 anemous 3 months ago

its pi not 2pi

 Shubham Singh 4 months ago
I m having problem in these two questions: 1) f(x)=x^3+6x^2+(9+2k)x+1 is increasing if k is 2)f(x)=cosx-2PX is monotonically decreasing for p:

HInt: 1. If we diffirenciate a cubic equation, we get the quadratic equation. Not for cubic polynomial to be increasing function, its derivative should be positive, i.e. quadratic polynomial...

 Vijay Mukati 2 months ago
area of triangle formed by pair of lines (x+2a)^2-3y^2 and x=a

The pair of lines (x+2a) 2 – 3y 2 = 0 passes through the origin ( just substitiute 0,0 and if it satisfies then it passes through the origin ). Now this pair of lines along with x=a forms a ...

 Ramvardhan one month ago
tan(c-d)

What exactly did you want as an answer? wanted me to derive the main equation. I think i misinterpreted your question.

 Neeti 2 months ago

yep i want not (a-b) i want (c-d)

 vivek 2 months ago

tan(a-b) = (tan A – tan B )/ 1+ tanA tanB

 Neeti 2 months ago
.If α,β are two values of θ satisfying equation cosθ/a+sinθ/b=1/c then.Prove that cot(α+β/2)=b/a. PLease kindly explain with full steps

cot( α+β/2) = (1+cos(α+β)) / (sin(α+β)) = [1+cosα*cosβ – sinα*sinβ] / [sinα*cosβ + cosα*sinβ] = [1+cos α*cos β*(1 – ta n α*tan β] / [cos α*cos β*(tan α + tan β) ] …...eq(1), take ur...

 Akshay 2 months ago

cot( α+β/2) = (1+cos(α+β)) / (sin(α+β)) = [1+cosα*cosβ – sinα*sinβ] / [sinα*cosβ + cosα*sinβ] = [1+cos α*cos β*(1 – ta n α*tan β] / [cos α*cos β*(tan α + tan β) ] …...eq(1), take ur...

 pa1 28 days ago
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