# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.9 | Set 1

### Solve the following differential equations:

### Question 1. x^{2}dy + y(x + y)dy = 0

**Solution:**

We have,

x

^{2}dy + y(x + y)dy = 0dy/dx = -y(x + y)/x

^{2}It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -vx(x + vx)/x

^{2}v + x(dv/dx) = -v – v

^{2}x(dv/dx) = -2v – v

^{2}On integrating both sides,

log|v/(v + 2)|

^{1/2 }= -log|x/c|v/(v + 2) = c

^{2}/x^{2}yx

^{2 }= (y + 2x)c^{2}(Where â€˜câ€™ is integration constant)

### Question 2. (dy/dx) = (y – x)/(y + x)

**Solution:**

We have,

(dy/dx) = (y – x)/(y + x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx – x)/(vx + x)

v + x(dv/dx) = (v – 1)/(v + 1)

x(dv/dx) = (v – 1)/(v + 1) – v

x(dv/dx) = (v – 1 – v

^{2 }– v)/(v + 1)x(dv/dx) = -(v

^{2 }+ 1)/(v + 1)On integrating both sides,

âˆ«vdv/(v

^{2}+1)+âˆ«dv/(v^{2}+1)=-âˆ«(dx/x)(1/2)log|v

^{2 }+ 1| + tan^{-1}(v) = log(c/x)log|(y

^{2 }+ x^{2})/x^{2}| + 2tan^{-1}(y/x) = log(c/x)^{2}log(y

^{2 }+ x^{2}) – log(x)^{2 }+ 2tan^{-1}(y/x) = log(c/x)^{2}log(y

^{2 }+ x^{2}) + 2tan^{-1}(y/x) = 2log(c) (Where â€˜câ€™ is integration constant)

### Question 3. (dy/dx) = (y^{2 }– x^{2})/2yx

**Solution:**

We have,

(dy/dx) = (y

^{2 }– x^{2})/2yxIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (v

^{2}x^{2 }– x^{2})/2vx^{2}v + x(dv/dx) = (v

^{2 }– 1)/2vx(dv/dx) = [(v

^{2 }– 1)/2v] – vx(dv/dx) = (v

^{2 }– 1 – 2v^{2})/2vx(dv/dx) = -(v

^{2 }+ 1)/2vOn integrating both sides,

log|v

^{2}+1| = -log(x) + log(c)log|v

^{2}+1| = log(c/x)y

^{2}/x^{2 }+ 1 = |c/x|(x

^{2 }+ y^{2}) = cx (Where â€˜câ€™ is integration constant)

### Question 4. x(dy/dx) = (x + y)

**Solution:**

We have,

x(dy/dx) = (x+y)

(dy/dx) = (x+y)/x

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + vx)/x

v + x(dv/dx) = (1 + v)

x(dv/dx) = 1

dv = (dx/x)

On integrating both sides,

âˆ«dv = âˆ«(dx/x)

v = log(x) + c

y/x = log(x) + c

y = xlog(x) + cx (Where â€˜câ€™ is integration constant)

### Question 5. (x^{2 }– y^{2})dx – 2xydy = 0

**Solution:**

We have,

(x

^{2 }– y^{2})dx – 2xydy = 0(dy/dx) = (x

^{2 }– y^{2})/2xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }– v^{2}x^{2})/2xvxv + x(dv/dx) = (1 – v

^{2})/2vx(dv/dx) = [(1 – v

^{2})/2v] – vx(dv/dx) = (1 – 3v

^{2})/2vOn integrating both sides,

-(1/3)log(1 – 3v

^{2}) = log(x) – log(c)log(1 – 3v

^{2}) = -log(x)^{3 }+ log(c)(x

^{2 }– 3y^{2})/x^{2 }= (c/x^{3})x(x

^{2 }– 3y^{2}) = c (Where â€˜câ€™ is integration constant)

### Question 6. (dy/dx) = (x + y)/(x – y)

**Solution:**

We have,

(dy/dx) = (x + y)/(x – y)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + vx)/(x – vx)

v + x(dv/dx) = (1 + v)/(1 – v)

x(dv/dx) = [(1 + v)/(1 – v)] – v

x(dv/dx) = (1 + v – v + v

^{2})/(1 – v)x(dv/dx) = (1 + v

^{2})/(1 – v)On integrating both sides,

âˆ«dv/(v

^{2 }+ 1) – âˆ«vdv/(v^{2 }+ 1) = âˆ«(dx/x)tan

^{-1}(v) – (1/2)log(v^{2 }+ 1) = log(x) + ctan

^{-1}(y/x) – (1/2)log(y^{2}/x^{2 }+ 1) = log(x) + ctan

^{-1}(y/x) – (1/2)log(y^{2 }+ x^{2}) + log(x) = log(x) + ctan

^{-1}(y/x) = (1/2)log(y^{2 }+ x^{2}) + c (Where â€˜câ€™ is integration constant)

### Question 7. 2xy(dy/dx) = (x^{2 }+ y^{2})

**Solution:**

We have,

2xy(dy/dx) = (x

^{2 }+ y^{2})(dy/dx) = (x

^{2 }+ y^{2})/2xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }+ v^{2}x^{2})/2xvxv + x(dv/dx) = (1 + v

^{2})/2vx(dv/dx) = [(1 + v

^{2})/2v] – vx(dv/dx) = (1 – v

^{2})/2vOn integrating both sides,

-log(1 – v

^{2}) = log(x) – log(c)log(1 – v

^{2}) = -log(x) + log(c)1 – y

^{2}/x^{2 }= (c/x)(x

^{2 }– y^{2}) = cx (Where â€˜câ€™ is integration constant)

### Question 8. x^{2}(dy/dx) = x^{2 }– 2y^{2 }+ xy

**Solution: **

We have,

x

^{2}(dy/dx) = x^{2 }– 2y^{2 }+ xy(dy/dx) = (x

^{2 }– 2y^{2 }+ xy)/x^{2}It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }– 2v^{2}x^{2 }+ xvx)/2xvxv + x(dv/dx) = (1 – 2v

^{2 }+ v)/x^{2}x(dv/dx) = (1 – 2v

^{2 }+ v) – vx(dv/dx) = (1 – 2v

^{2})dv/(1 – 2v

^{2}) = (dx/x)On integrating both sides,

dv/(1 – 2v

^{2}) = âˆ«(dx/x)(Where â€˜câ€™ is integration constant)

### Question 9. xy(dy/dx) = x^{2 }– y^{2}

**Solution:**

We have,

xy(dy/dx) = x

^{2 }– y^{2}(dy/dx) = (x

^{2 }– y^{2})/xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }– v^{2}x^{2})/xvxv + x(dv/dx) = (1 – v

^{2})/vx(dv/dx) = [(1 – v

^{2})/v] – vx(dv/dx) = (1 – 2v

^{2})/vvdv/(1 – 2v

^{2}) = (dx/x)On integrating both sides,

âˆ«vdv/(1 – 2v

^{2}) = âˆ«(dx/x)âˆ«4vdv/(1 – 2v

^{2}) = 4âˆ«(dx/x)-log(1 – 2v

^{2}) = 4log(x) – log(c)log(1 – 2v

^{2}) = log(c/x^{4})(1 – 2y

^{2}/x^{2}) = c/x^{4}(x

^{2}-2y^{2})/x^{2 }= c/x^{4}x

^{2}(x^{2 }– 2y^{2}) = c (Where â€˜câ€™ is integration constant)

### Question 10. ye^{x/y}dx = (xe^{x/y }+ y)dy

**Solution:**

We have,

ye

^{x/y}dx = (xe^{x/y }+ y)dy(dy/dx) = (xe

^{x/y }+ y)/ye^{x/y}It is a homogeneous equation,

So, put x = vy (i)

On differentiating both sides w.r.t x,

dx/dy = v + y(dv/dy)

So,

v + y(dv/dy) = (vye

^{vy/y }+ y)/ye^{vy/y}v + y(dv/dy) = (ve

^{v }+ 1)/e^{v}y(dv/dy) = [(ve

^{v }+ 1)/e^{v}] – vy(dv/dy) = (ve

^{v }+ 1 – ve^{v})/e^{v}y(dv/dy) = (1/e

^{v})e

^{v}dv = (dy/y)On integrating both sides,

âˆ«e

^{v}dv = âˆ«(dy/y)e

^{v }= log(y) + log(c)e

^{x/y }= log(y) + log(c) (Where â€˜câ€™ is integration constant)

### Question 11. x^{2}(dy/dx) = x^{2 }+ xy + y^{2}

**Solution:**

We have,

x

^{2}(dy/dx) = x^{2 }+ xy + y^{2}dy/dx = (x

^{2 }+ xy + y^{2})/x^{2}It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }+ xvx + v^{2}x^{2})/x^{2}v + x(dv/dx) = (1 + v + v

^{2})x(dv/dx) = (1 + v + v

^{2}) – vdv/(1 + v

^{2}) = (dx/x)On integrating both sides,

âˆ«dv/(1 + v

^{2}) = âˆ«(dx/x)tan

^{-1}(v) = log|x| + ctan

^{-1}(y/x) = log|x| + c (Where â€˜câ€™ is integration constant)

### Question 12. (y^{2 }– 2xy)dx = (x^{2 }– 2xy)dy

**Solution:**

We have,

(y

^{2 }– 2xy)dx = (x^{2 }– 2xy)dy(dy/dx) = (y

^{2 }– 2xy)/(x^{2 }– 2xy)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (v

^{2}x^{2 }– 2xvx)/(x^{2 }– 2xvx)v + x(dv/dx) = (v

^{2 }– 2v)/(1 – 2v)x(dv/dx) = [(v

^{2 }– 2v – v + 2v^{2})/(1 – 2v)]x(dv/dx) = 3(v

^{2 }– 1)/(1 – 2v)-(2v – 1)dv/(v

^{2 }– v) = 3(dx/x)On integrating both sides,

-âˆ«(2v – 1)dv/(v

^{2 }– v) = 3âˆ«(dx/x)-log|v

^{2 }– v| = 3log|x| – log|c|log|v

^{2 }– v| = log|c/x^{3}|(y

^{2}/x^{2 }– y/x) = (c/x^{3})(y

^{2 }– xy) = c/xx(y

^{2 }– xy) = c (Where â€˜câ€™ is integration constant)

### Question 13. 2xydx + (x^{2 }+ 2y^{2})dy = 0

**Solution:**

We have,

2xydx + (x

^{2 }+ 2y^{2})dy = 0dy/dx = -(2xy)/(x

^{2 }+ 2y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -(2xvx)/(x

^{2 }+ 2v^{2}x^{2})v + x(dv/dx) = -(2v)/(1 + 2v

^{2})x(dv/dx) = -[(2v)/(1 + 2v

^{2})] – vOn integrating both sides,

Substituting (3v + 2v

^{3}) = zOn differentiating both sides w.r.t x,

3(1 + 2v)dv = dz

(1 + 2v)dv = (dz/3)

(1/3)âˆ«(dz/z) = -âˆ«(dx/x)

(1/3)log|z| = -log|x| + log|c|

log|3v + 2v

^{3}| = log|c/x|^{3}3y/x + 2(y/x)

^{3 }= (c/x)^{3}(3yx

^{2 }+ 2y^{3}) = c (Where â€˜câ€™ is integration constant)

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