Hello Aspirant,

Hope you are doing well!!

It is quite difficult to make integration sign, In place of integration sign I'll put |, and in place of power I'll put **  when you are solve in your notebook you can place | sign to integration sign and ** to power sign.

Let X+Y = v

1 + dY/dX = dv/dX

dY/dX= dv/dX-1

dY/dX = sin (X + Y) + cos (X + Y)

dv/dX - 1 = sin v + cos v

dv / (1+cos v + sin v) = dX

Integrate both side

| dv / (1 + cos v + sin v ) = | dX | dv / ( 1 + ((1 - tan**2 (v/2)) / (1 + tan** 2 (v/2)) + ((2 tan (v/2) / (1 + tan **2 (v/2))

= |dX | sec**2 (v/2) dv / (2(1+ tan(v/2))

=|dx log (1 + tan (v/2)

= X+ c log ( 1+ tan (X + Y) / 2) = X + c

I hope this will help you.

Feel free to ask any query.

Hello,

This question is very simple. Divide the equation by x. logx, on both sides.

So we get :

dy/dx + y/ x.logx  = 2. logx / x .logx

i.e. dy/dx + y/ x logx = 2/x

Clearly this is a linear differential equation. So firstly we find the integrating factor.

I.F. = e ^ ( intg. { 1/ xlogx dx } ) ,

I.F. = logx

Now the solution of this differential equation is given by

y. I.F = intg. ( Q. I.F dx ) where Q = 2/x

Hence solution is : y. logx = (logx)^2 + c

Hope it helps.

hello aspirant,

you cannot understand differential equations without knowing or having knowledge on total integration.you need to have complete knowledge of both differentiation and integration to understand and solve problems on differential equations.

hope this helps,

thankyou.

Hello sakuntalaampolu,

Well, before I start solving it, I must admit it is going to be very lengthy as I am using bernoulli's homogeneous differential equation method:

Now, In first, if an equation is of form dy/dx = (ax + by + c)/(-bx+ay+f), then we should cross multiply and solve easily, But this is not the case here, So, put y = v + k, where v is variable and k is a constant, And put x = u + h, where u is variable and h is constant, and we get:

dy/dx = (3v-7u +3k-7h+7)/(3u-7v +3h-7k-3), now we know y = v + k => dy = dv and similarly du = dx So, dy/dx = dv/du and Now, lets make 3k-7h+7=0 and 3h-7k-3=0 because we want homogeneous equation and also k and h are constants while we will decide by solving above 2 equations having h and k , h=10/7 and k=1; Now we have dv/du = (3v-7u)/(3u-7v) Now, how do we solve homogeneous equation? => by putting v = ut where t is a variable, so dv/du = t + u.dt/du(you can also do dv/dt , No problem!) and we now get :

t + u.dt/du =  (3t-7)/(3-7t) => u.dt/du = 7(t^2 -1)/3-7t and rearranging we get:

((3-7t)/(7(t^2 -1))dt = du/u , Now Integrate both side, and you are going to get like : (-3/7 ln(t+1) -2/7 ln(t^2 -1) = ln u;( I hope you can solve it like 3-7t = (3-3t) -4t and rest you can do...)

And to get your answer, you have to put first t=v/u and then v = y-1(v = y-k and k=1) and u = x-10/7 and this can done by anyone,

And I am expecting you get the solution, Now Summarising:

first make homogenised equation by (using y=v+k and x=u+h) and after you get homogenised equation, always do like upper variable = t. lower variable like y = tx or v = ut and you can do the reverse also like u=vt, this will also give correct answer) and then you get a easily differentiable function to integrate and after integrating you should put the actual values or variables going in backward direction...

Good Luck!

Hope this helps, and feel free to ask any further query...

We are given that  dy/dt = 5y - 6t

rearranging which, we get, dy = 5y.dt - 6t.dt

Integrating both sides, we get

y = 5y.t - 3(t^2) + c

Given, y(0) = -2

which gives,

-2 = 0 + 0 +c

implies C = -2

Hence, the required equation is y = 5y.t - 3(t^2) - 2.

Hope it helps.

Hello Aspirant,

I'll recommend you to go through previous year paper for important questions and answers. Also you can purchase few books for sample question papers. These books covers almost all the important area of the subject. Solving these will help you with better understanding and practice.

Good Luck

Hello Prathik,

Hope everything going well.

Here is a method from which you can solve such equation:

Multiplying by 2x/x^2

2(xdx+ydy)+4y(xdy−ydx)/x^2=0

d(x^2+y^2)+4yd(y/x)=0

Converting to polar co ordinates

d(r^2)+4rsinθd(tanθ)=0

2rdr+4rsinθsec^2(θ)dθ=0

2rdr+4rsinθsec^2(θ)dθ=0

dr+2secθtanθdθ=0

Integrating,

r+2secθ=C

√( x^2+y^2)+2 √( 1+tan^2(θ))=C

√( x^2+y^2)+2 √( 1+(y^2/x^2))=C

(x+2) √(x^2+y^2)=Cx

Hope this helps you. Thank you.

All the best!!

Hi there!

Greetings!

Hope you are having a nice day!

Here I am giving you that notes of ordinary differential equation of IIT kharagpur :- http://www.math.iitb.ac.in/~siva/afs07.pdf so you can use it any I hope it might help you alot. So keep learning keep practicing and keep growing.

All the best for your future!

Thanks.

Hello Nirmal,

(2x+4y-6)dy=(2x-5y +3)dx

This is a very common type of  question that usually comes in the exams . This question is probably from the chapter of Differential Equations. As you understand that it is not that easy to solve the entire question here in the chat box so I have found an online solution of the equation that would help you.

Here's the link that will take you to the solution :-

https://math.stackexchange.com/questions/2001575/2x-5y3dx-2x4y-6dy/2001597#2001597

I hope it helped.

All the best.

Hey there

Maths and chemistry are scoring. Chemistry has no long writing SA type questions and thus you can score maximum in this subject. Physics has derivations and thus you have to learn stepwise skill to write derivations. Maths is calculative but write neat and clean do not cross too much in the question. Point wise writing is important.

Good luck