118) dy/dx + y/x = x2y3 or xdy + ydx = (xy)3dx or (xy)-3d(xy)=dx......i think nishant bhaiya u were luking 4 ds soln.......
A compilation of quesitons similar to the way bhargav had one :P (hence borrowed a part of the name..)
Keep solving and i will keep adding questions here...
To begin with first 5 are here...
Question 1) \left(\frac{dy}{dx}+1\right) x+\tan (x+y)=0
Question 2) f''(x)\left({1-f'(f(x))}\right) = f''(f(x)).(f'(x))^{2}
Question 3) xy' = y\ln(xy)
Question 4) y''=2x+(x^{2}-y')^{2}
Question 5) (1+x\sqrt{x^2+y^2})dx+(-1+\sqrt{x^2+y^2})ydy = 0
Question 6) x\frac{dy}{dx}+\left({\frac{dy}{dx}}\right)^2= y
Question 7) (x+y)(dx-dy)=dx+dy
Question 8) (2x^{2}+3y^{2}-7)xdx-(3x^{2}+2y^{2}-8)ydy = 0
Question 9) \frac{d^2y}{dx^2}(x^2+1) = 2x\frac{dy}{dx}
Question 10) \frac{dy}{dx}=\frac{yf'(x)-y^2}{f(x)}
-
UP 0 DOWN 0 7 63
63 Answers
11) (dy/dx + 1)x + tan(x + y) = 0 let x + y = z or 1 + dy/dx=dz/dx so we get xdz/dx + tanz = o or cotzdz = -dx/x
1
62for Q 13
why do you want to write it as teh last step..
leave it at dy/dx=c log y + A
and try and integrate from there!
62Question 21) \frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}
Question 22) (2x-y+1)dx+(2y-x-1)dy=0
Question 23) y^2dx+(x^2-xy)dy=0
Question 24) dy/dx=(\frac{x-y+3}{x-y+11})^2
Question 25) (y'-1/2y^2)sin x - y cos x = 0
Question 26) 4x^3y dx - (x^4+y^2)dy=0
Question 27)
17) dy / dx = (x + y - 1) / (x + y +1) ( using componendo and dividendo )
let x + y = a , so, 1 + dy/dx = da/dx , dy/dx = da/dx - 1
now,putting the value of dy/dx ,
da/dx = [ (a - 1) / (a+1) ] + 1 , da/dx = 2a / a+1
so, resolved as : ∫ da (a+1) / 2a = ∫ dx.
so, 1/2 ( a + lna ) = x + c.
values substituted cn be put 2 get the answer.
62@AIEEE there is an easier proof for 7 (more obvious proof than that) :P
117 ) .cn be written as : dy/dx + 2y / x = sinx / x
we know, if eqn, is of form : dy/dx + p.y = q , where p and q are functions of x , then:
y. e∫p.dx = ∫q . e∫p .dx dx
here , p = 2/x and q = sinx / x.
so,the integrating factor here is : e∫2/x dx = e2 lnx.
now , it cn be solved.
16) let z = dy/dx. so eqn. becomes y = xz + z2
so , by differentiating it , z = z + x dz/dx + 2z dz/dx
so , ( x + 2z ) dz/dx = 0. so, we get dz / dx = 0 i.e. z = constant and x + 2z = 0 i.e. z = - x/2.
first condition doesn't give any sufficient result. but by putting the 2nd condition ' we get :
4y + x2 = 0
19) let dy/dx = p. thus the equation becomes : dp / dx ( x2 + 1 ) = 2x p
so, ∫dp / p = ∫ dx ( 2x / x2 + 1 )
so , ln p + c = ln ( x2 + 1 )
so, pc = x2 + 1. now, putting p = dy/dx
so , c ∫dy = ∫ dx ( x2 + 1 )
now, it cn be solved.
2420
(x^2 \cos x -y)dx+xdy=0=>\frac {dy}{dx}=\frac {-x^2 \cos x +y} {x}=>\frac {dy}{dx}-\frac {y}{x}={-x\cos x}
=>y(\frac {1}{x})=\int \frac {1}{x}(-x\cos x) dx +C =>y=x[-\sin x +C]
2419\frac {dy}{dx}+\frac {y}{x}=\cos x
=> y(x)=\int x \cos x \ dx +C=>xy=x\sin x -(-\cos x) +C
=>xy=x\sin x +\cos x +C
2417
\frac {dy}{dx} +2\frac {y}{x}=\frac {\sin x }{x}
=> y .x^2=\int \ x^2 \frac {\sin x}{x} \ dx +C=>y .x^2=\int \ x \sin x \ dx
=> y .x^2=-x\cos x +\sin x +C
2415
\frac {dx}{dy}=\frac {3y}{x-z}=>(x-z)\ dx=3y\ dy ------------(1)
&\frac {dz}{dx}=-\frac {1}{3}=>3dz=-dx-----------(2)
3z=-x+C-------(3)
Putting values from (2) and (3) in (1)
=>-3(C-3z)\ dz=3y\ dy =>-(C-3z)dz=ydy =>-Cdz+3zdz=ydy
=>-Cz+\frac {3z^2 }{2}=\frac {y^2}{2} +C_1
Simialrly others
2413
y".y=c =>\frac {d^2y}{dx^2}y=c=>\frac {dy}{dx}=clogy+A;where \ A \ is\ a \ constant
Writing \ A=logB ;\ where \ B \ is \ another \ constant => \frac {dy}{dx}=log(y^c.B)
thinking ahead..its getting complex...
62Question 11) x\frac{dy}{dx}-y=3x^{2}y+xy^{2}
Question 12) xdy-ydx=(x^{2}+y^{2})dx
Question 13) y'' y = c
Question 14) \frac{\frac{d^2y}{dx^2}}{\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}^{\frac32}}=\frac1a\ (a>0)
Question 15) \large \begin{array}{ll}\frac{dx}{dt}=3y &\quad\\ \frac{dy}{dt}=x-z &\quad\\ \frac{dz}{dt}=-y &\quad\end{array}
Question 16) \frac{dx}{dt}-4x-\cos t=0
Question 17) x\frac{dy}{dx}+2y =\sin x
Question 18) \frac{dy}{dx}+\frac yx= x^2y^3
Question 19) y'+\frac{1}{x}y =\cos x
Question 20) (x^{2}\cos x-y)dx+xdy = 0
247
(x+y)(dx-dy)=dx+dy => \frac {dx+dy}{dx-dy}=x+y=>\frac {dy}{dx}=\frac {x+y-1}{x+y+1}
Put \ x+y=z =>1+\frac {dy}{dx}=\frac {dz}{dx}
=>\frac {dz}{dx}-1=\frac {z-1}{z+1} =>\frac {dz}{dx}=\frac {2z}{z+1}
\frac {z+1}{2z}dz=dx=>\frac {1}{2}[1+\frac {1}{z}]dz=dx=>\frac {1}{2}[z+logz]=x+C
62eureka.. ther is a 10 times simpler proof.. even aieee did the same in post 7! :)
3q1)x+y=t
ln(sinx)+lnx=c
q2)f''(f(x))f'(x)/1-f'(f(x))=d(f'(x))/f'(x) \Rightarrow ln(1-f'(f(x))-lnf'(x)+c
24oops i again spent my energy typing what had already been done[17]