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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Show that the differential equation `2y e^(x/y) dx+(y-2x e^(x y)`) `dy=0`ishomogeneous. Find the particular solution of this differential equation,given that `x=0`when `y=1.` |
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Answer» The given differential equation may be written as `(dx)/(dy)=(2xe^(x/y)-y)/(2ye^(x//y))`……………………(i) On dividing the Nr and Dr of RHS of (i) by y, we get `(dx)/(dy)={(2x/y.e^(x-y)-1)/(2e^(x//y))}=f(x/y)`…………….(ii) So, the given differential equation is homogeneous. Putting `x=vy` and `(dx)/(dy)=v+y(dv)/(dy)` in (ii), we get `v+y(dv)/(dy)=(2ve^(v)-1)/(2e^(v))` `rArr y(dv)/(dx)={(2ve^(v)-1)/(2e^(v))-v}=-1/(2e^(v))` `rArr 2e^(v)dv=-1/ydy` `rArr 2inte^(v)dy=-int1/ydy` [on integrating both sides] `rArr 2e^(v)=-log|y|+C` `rArr 2e^(x//y)+log|y|=C`...............(iii) `[therefore v=x/y]`. Putting, `y=1` and `x=0` in (iii), we get `C=2`. `therefore 2e^(x//y)+log|y|=2` is the required solution. |
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| 2. |
Find the particular solution, satisfying the givencondition, for the following differential equation:`(dy)/(dx)-y/x+cos e c (y/x)=0; y=0`when `x=1` |
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Answer» The given differential equation may be written as `(dy)/(dx)=y/x=-"cosec"y/x`………….(i) This is the form `(dy)/(dx)=f(y/x)`. So, it is homogeneous. Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=v-"cosec"v` `rArr -sinvdv=1/xdx` `rArr cosv=log|x|+C`, where C is an arbitary constant. `rArr cosy/x=log|x|+C...................(ii), [therefore v=y/x]`. Putting `x=1` and y=0 in (ii), we get C=1. Hence, `cosy/x=1+log|x|` is the required solution. |
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| 3. |
Find the particular solution of the differentialequation `(3x y+y^2)dx+(x^2=x y)dy=0;for x=1, y=1.` |
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Answer» We may write the given differential equation as `(dy)/(dx)=-(3xy+y^(2))/(x^(2)+xy)`……………(i) On dividing the Nr and Dr of RHS of (i) by `x^(2)`, we get `(dy)/(dx)=-{(3(y/x)+(y/x)^(2))/(1+y/x)}=f(y/x)`. So, the given differential equation is homogeneous. Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=-(3vx^(2)+v^(2)x^(2))/(x^(2)+vx^(2))=-(3v+v^(2))/(v+1)` `rArr x(dv)/(dx)={-(3v+v^(2))/(v+1)-v}=-(2(v^(2)+2v))/(v+1)` `rArr (v+1)/(v^(2)+2v)dv=-2/xdx` `rArr 1/2int(2(v+1))/(v^(2)+2v)dv+int2/xdx=log|C_(1)|` `rArr log|x^(2)sqrt(v^(2)+2v)|=log|C_(1)|` `rArr log|xsqrt(y^(2)+2xy)|=log|C_(1)| [therefore v=y/x]`. `rArr xsqrt(y^(2)+2xy)=+-|C_(1)|` `rArr x^(2)(y^(2)+2xy)=C,` Where `C=C_(1)^(2)`................(ii) `therefore x^(2)(y^(2)+2xy)=3` is the required solution. |
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| 4. |
Find the particular solution of the differential equation `(xe^(y//x)+y)dx=xdy`, given that `y(1)=0`. |
| Answer» Correct Answer - `log|x|+e^(-y//x)=1` | |
| 5. |
Find the particular solution of the differential equation `xe^(y/x)-y+xdy/dx=0,` given that `y(e)=0.` |
| Answer» Correct Answer - `y=-xlog(log|x|)` | |
| 6. |
`x(dy)/(dx)-y+xsin(y/x)=0` |
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Answer» Correct Answer - `xtan(y/(2x))=C` `(dy)/(dx)=y/x-sin(y/x)=f(y/x)`. `int(" cosec "vdv)=int1/xdx rArr log(log v)=logx + log C` `therefore log v =Cx = v=e^(Cx) rArr y=xe^(Cx)`. |
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| 7. |
Find the particular solution of the differential equation `(dy)/(dx)=(y(2y-x))/(x(2y+x))`,given that `y=1` when `x=1`. |
| Answer» Correct Answer - `y/x+1/2log|xy|=1` | |
| 8. |
Find the particular solution of the differential equation `{xsin^(2)(y/x)-y}dx+xdy=0`, it being given that `y=pi/4` when `x=1`. |
| Answer» Correct Answer - `coty/x=log|x|+1` | |
| 9. |
Solve : `(x+y)dy+(y-2x)dx=0` |
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Answer» Correct Answer - `y^(2)+2xy-2x^(2)=C` `int(v+1)/(v^(2)+2v-2)dv=-int(dx)/(x) rArr 1/2 int (2v+2)/(v^(2)+2v-2) dv= -int (dx)/x` `therefore 1/2log|v^(2)+2v-2|+log|x|=log|C_(1)|`. |
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| 10. |
`(x-y)dy-(x+y)dx=0` |
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Answer» Correct Answer - `tan^(-1)y/x=1/2log(x^(2)+y^(2))+C` `int(1-v)/(1+v^(2))dv=int1/xdx rArr int(dv)/(1+v^(2))-1/2int(2v)/(1+v^(2))dv=int1/xdx`. `therefore tan^(-1)v-1/2log(1+v^(2))=log|x|+C rArr tan^(-1)y/x=1/2log(x^(2)+y^(2))+C`. |
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| 11. |
`x^(2)dy+y(x+y)dx=0` |
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Answer» Correct Answer - `x^(2)y=C^(2)(y+2x)` `int(dv)/(v(v+2))=-int1/xdx` `rArr 1/2int{1/v-1/(v+2)}dv=-int1/xdx` [By Partial fractions] |
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| 12. |
Show that the differential equation `(dy)/(dx)=(y-x)/(y+x)` is homogenous and solve it. |
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Answer» Thegiven differential equation is `(dy)/(dx)=(y-x)/(y+x)`. …………….(i) On dividing the Nr and Dr of RHS of (i) by `x`, we get `(dy)/(dx)={(y/x-1)/(y/x+1)}=f(y/x)`. So, the given differential equtions is homogenous. Putting, `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=(vx-x)/(vx+x)` `rArr v+x(dv)/(dx)=(v-1)/(v+1)` `rArr x(dv)/(dx)=-(1+v^(2))/(1+v)` `rArr x(dv)/(dx)=-1/x dx` `rArr int(1+v)/(1+v^(2))dv=-int(dx)/x`. `rArr int1/(1+v^(2))dv+1/2int(2v)/(1+v^(2))dv=-int(dx)/x`. `rArr tan^(-)v+"log"|xsqrt(1+v^(2))|=C` `rArr tan^(-1)v+log|xsqrt(1+v^(2))|=C` `rArr tan^(-1)y/x+log|sqrt(x^(2)+y^(2))|=C` [Putting `v=y/x`] `rArr tan^(-1)y/x+1/2log(x^(2)+y^(2))=C`, which is the required solution. |
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| 13. |
Show that the differential equation is `x(dy)/(dx)-y=sqrt(x^(2)+y^(2))`, is homogenous and solve it. |
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Answer» The given differential equation may be written as `(dy)/(dx)=(y+sqrt(x^(2)+y^(2))/(x))`…………………(i) On dividing the Nr and Dr of RHS of (i) by `x`, we get So, the given differential equation is homogeneous. Putting, `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=(vx+sqrt(x^(2)+v^(2)x^(2))/(x))=v+sqrt(1+v^(2))` `rArr x(dv)/(dx)=sqrt(1+v^(2))` `rArr int(dv)/(1+v^(2))=int(dx)/x` `rArr log|v+sqrt(1+v^(2))|=log|C_(1)|` `rArr (v+sqrt(1+v^(2))/(x))=+-C_(1)=C` (say) `rArr v+sqrt(1+v^(2))=Cx` `rArr y+sqrt(x^(2)+y^(2))=Cx^(2)`, which is the required solution `[therefore v=yx]`. |
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| 14. |
Show that the differential equation `(xsqrt(x^(2)+y^(2)-y^(2)))dx+xydy=0` is homogenous and solve it. |
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Answer» The given differential equation may be written as `(dy)/(dx)=(y^(2)-xsqrt(x^(2)+y^(2))/(xy))`…………….(i) On dividing the Nr and Dr of RHS of (i) by `x^(2)`, we get `(dy)/(dx)={((y/x)^(2)-sqrt(1-(y/x)^(2)))/((y/x))}=f(y/x)`. So, the given differential equation is homogeneous. Putting, `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=(v^(2)x^(2)-xsqrt(x^(2)+v^(2)x^(2))/(vx^(2)))` `rArr x(dv)/(dx)=((v^(2)-sqrt(1+v^(2)))/(v)-v)` `rArr x(dv)/(dx)=-sqrt(1+v^(2))/(v)` `rArr intv/sqrt(1+v^(2))dy=-int(dx)/x` `rArr sqrt(1+v^(2))=-log|x|+C` `rArr sqrt(x^(2)+y^(2))+xlog|x|=Cx`, which is the required solution. |
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| 15. |
`(dy)/(dx)+(x-2y)/(2x-y)=0` |
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Answer» Correct Answer - `(y-x)=C(y+x)^(3)` `int(2-v)/(v^(2)-1)dv=int1/xdx`. `(2-v)/(v^(2)-1)={1/(2(v-1))-3/(2(v+1))}` [by partial fractions]. |
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| 16. |
`x^(2)(dy)/(dx)+y^(2)=xy` |
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Answer» Correct Answer - `Cx=e^(x//y)` `x(dv)/(dx)=-v^(2) rArr -1/v^(2)dv=int-1/xdx rArr 1/v =logx +log C`. `therefore logCx=x/y rArr Cx=e^(x//y)` |
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| 17. |
`y^(2)+(x^(2)-xy)(dy)/(dx)=0` |
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Answer» Correct Answer - `y=x{C+log|y|}` `int(v-1)/(v(v+1))dv=int1/xdx rArr int{2/(v+1)-1/v} dv=1/xdx.` |
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| 18. |
`x(dy)/(dx)-y=2sqrt(y^2-x^2)` |
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Answer» Correct Answer - `y+sqrt(y^(2)-x^(2))=C|x|^(3)` `int1/(sqrt(v^(2)-1))dv=int2/xdx rArr log|v+sqrt(v^(2)-1)|=2log|x|+log|C_(1)|`. |
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| 19. |
`y^2dx+(x^2+x y+y^2)dy=0` |
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Answer» Correct Answer - `log|y/(y+x)|+log|x|+x/(y+x)=C` `int1/(v(v+1)^(2))dv=-int1/xdx`. Let `1/(v(v+1))^(2)=A/v+B/(v+1)+C/(v+1)^(2)` This gives A=1, B`=-1` and `C=-1`. |
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| 20. |
Show that the differential equation `x^(2)(dy)/(dx)=(x^(2)-2y^(2)+xy)` is homogenous and solve it. |
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Answer» The given differential equation may be written as `(dy)/(dx)=(x^(2)-2y^(2)+xy)/(x^(2))`. On dividing the Nr and Dr of RHS of (i)by `x^(2)`, we get `(dy)/(dx)={1-2(y/x)^(2)+(y/x)}=f(y/x)`. So, the given differential equation is homogenous. Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=(x^(2)-2v^(2)x^(2)+vx^(2))/(x^(2))` `rArr v+x(dv)/(dx)=1-2v^(2)+v` `rArr x(dv)/(dx)=1-2v^(2)` `rArr int (dv)/(1-2v^(2))=int(dx)/x`. `rArr 1/2int(dv)/(1-2v^(2))=int(dx)/x`. `rArr 1/2int(dv)/({(1/sqrt(2))^(2)-v^(2)})=int(dx)/x`. `rArr 1/2.1/(2 xx 1/sqrt(2))log|(1/sqrt(2)+v)/(1/sqrt(2)-v)|=log|x|+C` `rArr 1/(2sqrt(2))log|(x+sqrt(2)y)/(x-sqrt(2)y)|-log|x|=C [therefore v=y/x]`. This is the required solution. |
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| 21. |
`x^(2)(dy)/(dx)=x^(2)+xy+y^(2)` |
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Answer» Correct Answer - `tan^(-1)y/x=log|x|+C` `int 1/(1+v^(2))dv=int1/xdx rArr tan^(-1)v=log|x|+C`. |
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| 22. |
The slope of the tangent to a curve at any point `(x ,y)`on its given by `y/x-coty/xdotcosy/x ,(x >0,y >0)`and the curve passes though the point `(1,pi//4)dot`Find the equation of the curve. |
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Answer» Correct Answer - `secy/x+log|x|=sqrt(2)` `(dy)/(dx)={y/x-cot y/x cosy/x}=f(y/x)`, whch is homogenous. Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)`, we get `int sec v tan v dv =-int(dx)/(x) rArr sec v=-log|x|+C` `therefore sec(y/x)+log|x|=C`. Putting `x=1` and `y=pi/4`, we get `C=sqrt(2)`. |
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| 23. |
`(dy)/(dx)=(2xy)/(x^(2)-y^(2))` |
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Answer» Correct Answer - `y=x{C+log|y|}` `x(dv)/(dx)=(v+v^(3))/(1-v^(2)) rArr int (1-v^(2))/(v(1+v^(2))dv)=int1/xdx` `rArr int1/v-(2v)/(1+v^(2))dv=-1/xdx`. [By partial fractions]. |
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| 24. |
`(dy)/(dx)=(x^(2)+y^(2))/(2xy)` |
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Answer» Correct Answer - `(x^(2)-y^(2))=Cx` `x(dv)/(dx)=(1-v^(2))/(2v) rArr int (-2v)/(1-v^(2))dv=-int1/xdx` `rArr log|1-v^(2)|+log|x|=log|C_(1)|`. |
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| 25. |
Solve the following differential equations:`(x^2+3x y+y^2^)dx-x^2dy=0` |
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Answer» Correct Answer - `log|x|=x/(y+x)=C` `x(dv)/(dx)=(1+2v+v^(2))=(v+1)^(2) rArr int(dv)/(v+1)^(2)=int 1/x dx rArr log|x|-1/(v+1)=C` |
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| 26. |
`(x-y)dy/dx=x+3y` |
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Answer» Correct Answer - `log|x+y|+(2x)/(x+y)=C` `x(dv)/(dx)=(1+v)^(2)/(1-v) rArr int (v-1)/(v+1)^(2)dx=-int(dx)/x`. `rArr int{((v+1)-2)/(v+1)^(2)}dv=-int(dx)/x rArr int1/(v+1)^(2)dv+int(dx)/x=C` |
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| 27. |
`(x-sqrt(xy))dy=ydx` |
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Answer» Correct Answer - `2sqrt(x/y)+log|y|=C` `int(1-sqrt(v))/(v^(3//2))dv=int1/xdx rArr int{1/(v^(3//2)-sqrt(v)/(v^(3//2)))}dv=int 1/x dx` `rArr int v^(-3//2)dv-int1/vdv=log|x|-C rArr -2/sqrt(v)-log|v|=log|x|-C` `rArr 2sqrt(x/y)+log|y|=C`. |
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| 28. |
`(x^3+3x y^2)dx+(y^3+3x^2y)dy=0` |
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Answer» Correct Answer - `y^(4)+6x^(2)y^(2)+x^(2)=C` `int(v^(3)+3v)/(v^(4)+6v^(2)+1)dv=-int1/xdx` `rArr 1/4int(4v^(3)+12v)/(v^(4)+6v^(2)+1)dv=-int1/xdx` `rArr 1/4log|v^(4)+6v^(2)+1|+log|x|=log|C_(1)|` `rArr (v^(4)+6v^(2)+1)x^(4)=C_(1)^(4)=C` (say) `rArr y^(4)+6x^(2)y^(2)+x^(4)`=C. |
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| 29. |
`2xydx+(x^(2)+2y^(2))dy=0` |
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Answer» Correct Answer - `3x^(2)y+2y^(3)=C` `x(dv)/(dx) = -(3v+2v^(3))(1+2v^(2)) rArr int(1+2v^(2))/(3v+2v^(3))dv+int(dx)/(x)=log|C_(1)|`. `therefore 1/3int(3+6v^(2))/(3v+2v^(3))dv+log|x|=log|C_(1)|` `rArr log|3v+2v^(3)|+log|x^(3)|=log|C_(1)^(3)|`. |
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| 30. |
Show that the differential equation `(y^2-x^2)dy=3xy dx` is homogenous and solve it |
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Answer» The given differential equation may be written as `(dy)/(dx)=(3xy)/(y^(2)-x^(2))`…………..(i) On dividing the Nr and Dr of RHS of (i) by `x^(2)`, we get `(dy)/(dx)=(3(y/x))/{(y/x)^(2)-1}=f(y/x)`. Thus, the given differential equation is homogenous. Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=(3v)/(v^(2)-1)` `rArr x(dv)/(dx){(3v)/(v^(2)-1)-v}=(4v-v^(3))/(v^(2)-1)` `rArr (v^(2)-1)/(4v-v^(3))dv=1/xdx` `rArr int (v^(2)-1)/(v(2-v)(2+v))dv=1/xdx`...........(ii) Let `(v^(2)-1)/(v(2-v)(2+v))=A/v+B/(2-v)+C/(2+v)`. Then, `(v^(2)-1)-= A(2-v)(2+v)+Bv(2+v)+Cv(2-v)`.............(iii) Putting, `v=0` on each side of (iii), we get `A=-1/4`. Putting v=2 on each side of (iii), we get `B=3/8`. Putting `v=-2` on each side of (iii), we get `C=-3/8`. `therefore (v^(2)-1)/(v(2-v)(2+v))=-1/(4v)+3/(8(2-v))-3/(8(2+v))`................(iv) Putting, these values from (iv) in (ii), we get `-1/4int(dv)/(v)=3/8int(dv)/(2+v)=int1/xdx` `rArr int1/xdx+1/4int(dv)/v+3/8int(-dv)/(2-v)=int1/xdx` `rArr int1/xdx+1/4int(dv)/v+3/8int(-dv)/(2-v)+3/8int(dv)/(2+v)=log|C_(1)|` `rArr 8log|x|+2log|v|+3log|2-v|+3log|2+v|=8log|C_(1)|` `rArr |x^(8)v^(2)(2-v)^(3)(2+v)^(3)|=C_(1)^(8)=C` (say) `rArr |x^(8)v^(2)(2-v)^(3)(2+v)^(3)|=C_(1)^(8)=C` (say) `rArr x^(8)y^(2)/x^(2)(2-y/x)^(3)(2+y/x)^(3)=C` `rArr y^(2)(2x-y)^(3)(2x+y)^(3)=C`, where C is an arbirary constant. `rArr y^(2)(4x^(2)-y^(2))^(2)=C`, which is the required solution. |
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| 31. |
Show that the differential equation `y^(2)dx+(x^(2)-xy+y^(2))dy=0` is homogenous and solve it. |
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Answer» The given differential equation may be written as `(dy)/(dx)=-y^(2)/(x^(2)-xy+y^(2))`……………..(i) On dividing the Nr and Dr of RHS of (i) by `x^(2)`, we get `(dy)/(dx)=-(y/x)^(2)/(1-y/x+(y/x)^(2))=f(y/x)`. Thus, the given differential equation is homogeneous. Putting, `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx) =(-v^(2))(1-v+v^(2))` `rArr x(dv)/(dx)-v^(2)/(1-v+v^(2)+v]` `rArr x(dv)/(dx)=-(v+v^(3))/(1-v+v^(2))` `rArr (1-v+v^(2))/(v(1+v^(2)))dv=-1/xdx` `rArr int((1+v^(2))-v)/(v(1+v^(2)))dv=-int(dx)/x`. `rArr int(dv)/v-int(dv)/(1+v^(2))+int(dx)/x=logC` `rArr log|v|-tan^(-1)v+log|x|=logC` `rArr tan^(-1)v=log(|vx|)/C` `rArr tan^(-1)(y/x)=log(|y|/C) [therefore v=y/x]` `rArr |y|/c=e^(tan^(-1)(y//x))` `rArr |y|=Ce^(tan^(-1)(y//x))`, which is the required solutions. |
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| 32. |
Show that the differential equation `x(dy)/(dx)=y-x tan(y/x)` is homogenous and solve it. |
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Answer» The given differential equation may be written as `(dy)/(dx)=y/x-tany/x`…………….(i) This is of the form, `(dy)/(dx)=f(y/x)`. So,the given differential equation is homogeneous. Putting, `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `v+x(dv)/(dx)=v-tanv` `rArr x(dv)/(dx)=-tanv` `rArr (dv)/(tanv)=-(dx)/x` `rArr |
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| 33. |
Show that the differential equation `(x^(3)-3xy^(2))dx=(y^(3)-3x^(2)y)dy` is homogenous and solve it. |
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Answer» The given differential equation may be written as `(dy)/(dx)=(x^(3)-3xy^(2))/(y^(3)-3x^(2)y)`………..(i) On dividing the Nr and Dr of RHS of (i) by `x^(3)`, we get `(dy)/(dx)=(1-3(y/x)^(2))/((y/x)^(2)-3(y/x))=f(y/x)`. Thus, the given differential equation is homogenous. Putting, `v=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `(dy)/(dx)=(1-3(y/x)^(2))/((y/x)^(3)-3(y/x))=f(y/x)`. Thus, the given differential equation is homogeneous. Putting, `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get `=v+x(dv)/(dx)=(x^(3)-3v^(2)x^(3))/(v^(3)x^(3)-3vx^(3))` `rArr v+x(dv)/(dx)=(1-3v^(2))/(v^(3)-3v)` `rArr x(dv)/(dx)=((1-3v^(2))/(v^(3)-3v)-v)` `rArr x(dv)/(dx)=(1-v^(4))/(v^(3)-3v)` `rArr (3v-v^(3))/(v^(4)-1)dv=int(dx)/x`.....................(ii) Let `((3v-v^(3))/(v^(4)-1)=A/(v-1)+B/(v+1)+(Cv+D)/(v^(2)+1))`. Then, `(3v-v^(3))-=A(v+1)(v^(2)+1)+B(v-1)(v^(2)+1)+(Cv+D)(v-1)(v+1))`............(iii) Putting v=1 on each side of (iii), we get `A=1/2`. Putting v`=-1` on each side of (iii), we get `B=1/2`. Comparing coefficients of `v^(3)` on both sides of (iii), we get `A+B+C=-1 rArr 1/2+1/2+C=-1 rArr C=-2`. Comparing the independent terms on both sides of (iii), we get `A-B-D=0 rArr D=(A-B)=(1/2-1/2)=0`. `therefore ((3v-v^(3))/(v^(4)-1)=1/(2(v-1))+1/(2(v+1))-(2v)/(v^(2)+1))`. Putting this value in (ii), we get `1/2int(dv)/(v-1)+1/2int(dv)/(v+1)-int(2v)/(v^(2)+1)dv=int(dx)/x` `rArr 1/2log|v-1|+1/2log|v+1|-log|v^(2)+1|=log|x|+log|C|`, where C is an arbitrary constant. `rArr log|v-1|+log|v+1|-2log|v^(2)+1|=2log|x|+2log|C|` `rArr log|((v-1)(v+1))(v^(2)+1)^(2)|=log|C^(2)x^(2)| rArr log|(v^(2)-1)/(v^(2)+1)^(2)|=log|C^(2)x^(2)|` `rArr (y^(2)-x^(2))=C^(2)(y^(2)+x^(2))^(2) [therefore v=y/x]`. Hence, `(y^(2)-x^(2))=C^(2)(y^(2)+x^(2))^(2)` is the required solutions. |
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| 34. |
Solve - `x^2dy/dx = 2xy + y^2` |
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Answer» Correct Answer - `y=Cx(x+y)` `int1/(v(1+v))dv=int1/xdx rArr int {1/v-1/(1+v)}dv=int1/x dx` [ by partial fractions]. |
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