29For the first one...substitute x + y = t...
in the third substitute 4x + y + 1 = t..
for the fourth one on observation it looks like the equation is x4y3 = -b2/a2 xy + c...par LHS main coefficients stidfy nahi ho rahey.. : (
1) cos (x+y) dy = dx
2) x+ydydxy-xdydx = (1 + x2y2) ( ln ( x2+y2 ))
3) dydx = (4x + y + 1)2
4) 2x3y3 + x4y2 dydx = -b2a2y - b2a2 x dydx
[ GUYS POST THE SOLN. FULL AS I CANNOT CONFIRM UR ANS. SINCE I DON'T HAVE THEM ]
-
UP 0 DOWN 0 1 11
11 Answers
2929cost = dx/dy ....
x + y = t
diff. wrt to y
dx/dy + 1 = dt/dy
cost + 1 = dt/ dy..
now integrate it...write cost in terms of tant/2
cost = 1-tan2t/2 / 1+tan2t/2
13) dtdx = t2 + 4.
dtt2 + 4 = dx.
x = 12 tan -1 (t2)
x = 12 tan -1 (4x + y + 12)
WAT ABT THE 2ND ONE????????
13Q2) \ My \ start...\\ y/x =t => y' = x.t'+t\\ \\ \left[ 1+ t \left(x.\frac{dt}{dx}+t\right)\right] = \left(-x.\frac{dt}{dx}}\right)\left(1+ \frac{1}{t}\right)\left[2ln(x)+ ln(1+t^2)\right]
Which reduces to....
\frac{dt}{dx}\left[ t+ ln(1+t^2)^{1+1/t} + ln(x)^{2/t +2}\right] + \frac{(1+t^2)}{x} = 0
Now, if this can be converted to a linear one; ...can't see ny phurther....
14) Divide whole equation by x2y2 and multiply by dx
2xydx+x^{2}dy=-\frac{b^2}{a^2}\frac{ydx+xdy}{x^2y^2}
yd(x^2)+x^{2}dy=-\frac{b^2}{a^2}\frac{d(xy)}{x^2y^2}
d(x^2y)=-\frac{b^2}{a^2}\frac{d(xy)}{x^2y^2}
integrate :
x^2y^2-cxy=\frac{b^2}{a^2}
12)
\frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=\left(1+\frac{x^2}{y^2} \right)ln(x^2+y^2) \\ \\\frac{xdx+ydy}{ydx-xdy}=\left(1+\frac{x^2}{y^2} \right)ln(x^2+y^2) \\ \\\frac{xdx+ydy}{\frac{ydx-xdy}{y^2}}=\left(x^2+y^2 \right)ln(x^2+y^2) \\ \\\frac{d(x^2+y^2)}{\left(x^2+y^2 \right)ln(x^2+y^2)}=d\left(\frac{x}{y} \right)
Now take:
x^2+y^2=z \\\\and,\: \frac{x}{y}=t \\\\equn.\: becomes: \frac{dz}{zlnz}=dt \\\\or,\: ln(ln(z))=t+c \\\\solution:\: ln(ln(x^2+y^2))=\frac{x}{y}+c \\\\or,\:ln(x^2+y^2)=ke^{\frac{x}{y}}