Solving Homogeneous Differential Equations
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Q.
Solution of the differential equation √xdx+√ydy√xdx−√ydy=√y3x3 is given by
32log(yx)+log∣∣∣x3/2+y3/2x3/2∣∣∣+tan−1(yx)3/2+c=0
23log(yx)+log∣∣∣x3/2+y3/2x3/2∣∣∣+tan−1yx+c=0
23log(yx)+log(x+yx)+tan−1(y3/2x3/2)+c=0
None of the above
Q.
The equation of the normal of the curve at is
Q. An equation of the curve satisfying xdy−ydx=√x2−y2 dx and y(1)=0 is
- y=x2log|sinx|
- y=xsin(log|x|)
- y2=x(x−1)2
- y=2x2(x−1)
Q. The family of curves which satisfies the differential equation y exydx=(x exy+y2)dy, (y≠0) is
(where ′C′ is the constant of integration )
(where ′C′ is the constant of integration )
- y exy=C
- y+exy=C
- x−eyx=C
- y−exy=C
Q.
The general solution of y2 dx+(x2−xy+y2)dy=0 is
[EAMCET 2003]
tan−1(xy)+log y+c=0
2 tan−1(xy)+log x+c=0
log(y+√x2+y2)+log y+c=0
sin h−1(xy)+log y+c=0
Q. The particular solution of the differential equation xdydx=yln(yx), x≠0 is y=f(x). If f(1)=e2, then
- f(x)=xe1+x2
- f(x)=xe1+x
- there are more than one tangent parallel to x−axis to the curve y=f(x)
- there is exactly one tangent parallel to x−axis to the curve y=f(x)
Q.
What is the value of in the equation ?
Q. Solution of the differential equation : dydx=3x2y4+2xyx2−2x3y3 is
- x2y2+x2y=c
- x3y2+x2y=c
- x3y2+y2x=c
- x2y3+x2y=c
Q. The solution of x2dydx−xy=1+cosyx is
- tany2x=C−12x2
- tanyx=C+1x
- cos(yx)=1+Cx
- x2=(C+x2)tanyx
Q. The curve amongst the family of curves represented by the differential equation, (x2−y2)dx+2xy dy=0 which passes through (1, 1), is :
- a circle with centre on the x-axis
- a circle with centre on the y-axis
- an ellipse with major axis along the y-axis
- a hyperbola with transverse axis along the x-axis
Q. The solution of (y+x+5)dy=(y−x+1)dx is
- log((y+3)2+(x+2)2)+tan−1y+3x+2=C
- log((y+3)2+(x−2)2)+tan−1y−3x−2=C
- log((y+3)2+(x+2)2)+2tan−1y+3x+2=C
- log((y+3)2+(x+2)2)−2tan−1y+3x+2=C
Q. An equation of the curve satisfying xdy−ydx=√x2−y2 dx and y(1)=0 is
- y=x2log|sinx|
- y=xsin(log|x|)
- y2=x(x−1)2
- y=2x2(x−1)
Q. The general solution of y2 dx+(x2−xy+y2)dy=0 is
- tan−1(xy)+log y+c=0
- 2tan−1(xy)+log x+c=0
- log(y+√x2+y2)+log y+c=0
- sin h−1(xy)+log y+c=0
Q. The solution of (y+x+5)dy=(y−x+1)dx is
- log((y+3)2+(x+2)2)+tan−1y+3x+2=C
- log((y+3)2+(x−2)2)+tan−1y−3x−2=C
- log((y+3)2+(x+2)2)+2tan−1y+3x+2=C
- log((y+3)2+(x+2)2)−2tan−1y+3x+2=C
Q. If xdydx=y(log y−log x+1), then the solution of the equation is
- y log(xy)=cx
- x log(yx)=cy
- log(yx)=cx
- log(xy)=cy