第12类RD Sharma解–第22章微分方程–练习22.9 |套装3

问题27.(x ² – 2xy)dy +(x ² – 3xy + 2y ² )dx = 0

解决方案：

We have,

(x² – 2xy)dy + (x² – 3xy + 2y²)dx = 0

(dy/dx) = (x² – 3xy + 2y²)/(2xy – y²)

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) = (x² – 3xvx + 2v²x²)/(2xvx – x²)

v + x(dv/dx) = (1 – 3v + 2v²)/(2v – 1)

x(dv/dx) = [(1 – 3v + 2v²)/(2v – 1)] – v

x(dv/dx) = (1 – 3v + 2v² – 2v² + v)/(2v – 1)

x(dv/dx) = (1 – 2v)/(2v – 1)

x(dv/dx) = -1

dv = -(dx/x)

On integrating both sides,

∫dv = -∫(dx/x)

v = -log|x| + log|c|

(y/x) + log|x| = log|c|  (Where ‘c’ is integration constant)

问题28. x(dy / dx)= y – xcos ² (y / x)

解决方案：

We have,

x(dy/dx) = y – xcos²(y/x)

(dy/dx) = y/x – cos²(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) = v – cos²(v)

x(dv/dx) = -cos²(v)

dv/cos²(v) = -(dx/x)

On integrating both sides,

∫dv/cos²(v) = -∫(dx/x)

∫sec²vdv = -∫(dx/x)

tan(v) = -log|x| + log|c|

tan(y/x) = log|c/x|  (Where ‘c’ is integration constant)

问题29. x(dy / dx)– y =2√(y ² – x ² )

解决方案：

We have,

x(dy/dx) – y = 2√(y² – x²)

(dy/dx) = [2√(y² – x²) + 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,

$v+x\frac{dv}{dx}=\frac{2\sqrt{v^2x^2-x^2}+vx}{x}$

$v+x\frac{dv}{dx}=2\sqrt{v^2-1}+v$

x(dv/dx) = 2√(v² – 1)

dv/√(v² – 1) = 2(dx/x)

On integrating both sides,

∫dv/√(v² – 1) = 2∫(dx/x)

log|v + √(v²– 1)| = 2log(x) + log(c)

|v + √(v² – 1)| = |cx²|

$\frac{y}{x}+\sqrt{\frac{y^2}{x^2}-1}=|cx^2|$

$y+\sqrt{y^2-x^2}=cx^3$  (Where ‘c’ is integration constant)

问题30. xcos(y / x)(ydx + xdy)= ysin(y / x)(xdy – ydx)

解决方案：

We have,

xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

xycos(y/x)dx + x²cos(y/x)dy = xysin(y/x)dy – y²sin(y/x)dx

x²cos(y/x)dy – xysin(y/x)dy = -y²sin(y/x)dx – xycos(y/x)dx

$\frac{dy}{dx}=\frac{xysin(\frac{y}{x})+y^2cos(\frac{y}{x})}{xysin(\frac{y}{x})-x^2cos(\frac{y}{x})}$

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,

$v+x\frac{dv}{dx}=\frac{xvxsinv+v^2x^2cosv}{xvxsinv-x^2cosv}$

v + x(dv/dx) = (vcosv + v²sinv)/(vsinv – cosv)

x(dv/dx) = [(vcosv + v²sinv)/(vsinv – cosv)] – v

x(dv/dx) = (vcosv + v²sinv – v²sinv + vcosv)/(vsinv – cosv)

x(dv/dx) = (2vcosv)/(vsinv – cosv)

[(vsinv – cosv)/(vcosv)]dv = 2(dx/x)

On integrating both sides,

∫tanvdv – ∫(dv/v) = 2log|x| + log|c|

log|secv| – log|v| = log|cx²|

log|(secv/v)| = log|cx²|

(x/y)sec(y/x) = cx²

sec(y/x) = cxy (Where ‘c’ is integration constant)

问题31.(x ² + 3xy + y ² )dx – x ² dy = 0

解决方案：

We have,

(x² + 3xy + y²)dx – x²dy = 0

dy/dx = (x² + 3xy + y²)/x²

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) = (x² + 3xvx + v²x²)/x²

v + x(dv/dx) = (1 + 3v + v²)

x(dv/dx) = (1 + 3v + v²) – v

x(dv/dx) = (1 + 2v + v²)

x(dv/dx) = (1 + v)²

dv/(1 + v)² = (dx/x)

On integrating both sides,

∫dv/(1 + v)² = ∫(dx/x)

-[1/(v + 1)] = log|x| – c

$-\frac{1}{\frac{y}{x}+1}=log|x|-c$

x/(x + y) + log|x| = c    (Where ‘c’ is an integration constant)

问题32.(x – y)(dy / dx)=(x + 2y)

解决方案：

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(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 + 2vx)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v²)/(1 – v)

x(dv/dx) = (1 + v + v²)/(1 – v)

(1 – v)dv/(1 + v + v²) = (dx/x)

On integrating both sides,

∫[(1 – v)/(1 + v + v²)]dv = ∫(dx/x)

$-\frac{1}{2}∫\frac{2v-2}{v^2+v+1}dv=∫\frac{dx}{x}$

$∫\frac{(2v+1)-3}{v^2+v+1}dv=-2∫\frac{dx}{x}$

$∫\frac{2v+1}{v^2+v+1}dv-∫\frac{3dv}{v^2+2v(\frac{1}{2})+(\frac{1}{2})^2-(\frac{1}{2})^2+1}=-2log|x|+c$

$log|v^2+v+1|-∫\frac{3dv}{(v+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=-log|x^2|+c$

$log|\frac{y^2}{x^2}+\frac{y}{x}+1|-3(\frac{2}{\sqrt{3}})tan^{-1}(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}})+log|x^2|=c$

$log|y^2+xy+x^2|=2\sqrt{3}tan^{-1}(\frac{2y+x}{x\sqrt{3}})+c$  (Where ‘c’ is an integration constant)

问题33.(2x ² y + y ³ )dx +(xy ² – 3x ² )dy = 0

解决方案：

We have,

(2x²y + y³)dx + (xy² – 3x²)dy = 0

dy/dx = (2x²y + y³)/(3x³ – xy²)

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) = (2x²vx + v³x³)/(3x³ – xv²x²)

v + x(dv/dx) = (2v + v³)/(3 – v³)

x(dv/dx) = [(2v + v³)/(3 – v³)] – v

x(dv/dx) = (2v + v³ – 3v + v³)/(3 – v³)

(3 – v³)dv/(2v³ – v) = (dx/x)

On integrating both sides,

∫[(3 – v³)/(2v³ – v)]dv = ∫(dx/x)

$∫\frac{3-v^2}{v(2v^2-1)}dv=∫\frac{dx}{x}$

Using partial fraction,

$\frac{3-v^2}{v(2v^2-1)}=\frac{A}{v}+\frac{Bv+C}{2v^2-1}$

3 – v² = A(2v² – 1) + (Bv + C)v

3 – v² = 2Av² – A + Bv² + Cv

3 – v² = v²(2A + B) + Cv – A

On comparing the coefficients, we get

A = -3,

B = 5,

C = 0,

$-3∫\frac{dv}{v}+∫\frac{5v}{2v^2-1}dv=∫\frac{dx}{x}$

$-3∫\frac{dv}{v}+\frac{5}{4}∫\frac{4v}{2v^2-1}dv=∫\frac{dx}{x}$

-3log|v|+(5/4)log|2v²-1|=log|x|+log|c|

-12log|v|+5log|2v²-1|=4log|x|+4log|c|

$log|\frac{2v^2-1}{v^{12}}|=log|x^4c^4|$

$(2\frac{y^2}{x^2}-1)^5=x^4c^4\frac{y^{12}}{x^{12}}$

$\frac{2y^2-x^2}{x^{10}}=x^4c^4\frac{y^{12}}{x^{12}}$

|2y² – x²|⁵ = x²c⁴y¹²  (Where ‘c’ is an integration constant)

问题34. x(dy / dx)– y + xsin(y / x)= 0

解决方案：

We have,

x(dy/dx) – y + xsin(y/x) = 0
x(dy/dx) = y – xsin(y/x)

(dy/dx) = [y – xsin(y/x)]/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 – xsinv]/x

v + x(dv/dx) = (v – sinv)

x(dv/dx) = -sinv

cosecvdv = -(dx/x)

On integrating both sides,

∫cosecvdv = -∫(dx/x)

-log|cosecv + cotv| = -log|x| + log|c|

-log|(1/sinv) + (cosv/sinv)| = -log|x/c|

|(1 + cosv)/sinv| = |x/c|

xsinv = c(1 + cosv)

xsin(y/x) = c[1 + cos(y/x)]    (Where ‘c’ is integration constant)

问题35。ydx + {xlog(y / x)} dy – 2xydy = 0

解决方案：

We have,

ydx + {xlog(y/x)}dy – 2xydy = 0

y + {xlog(y/x)}(dy/dx) – 2xy = 0

$\frac{dy}{dx}=\frac{y}{2x-xlog(\frac{y}{x})}$

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,

$v+x\frac{dv}{dx}=-\frac{vx}{2x-xlogv}$

v + x(dv/dx) = v/(2 – logv)

x(dv/dx) = [v/(2 – logv)] – v

x(dv/dx) = (v – 2v + vlogv)/(2 – logv)

x(dv/dx) = -v(logv – 1)/(logv – 2)

$\frac{(logv-2)}{v(logv-1)}dv=-\frac{dx}{x}$

On integrating both sides,

$∫\frac{(logv-2)}{v(logv-1)}dv=-∫\frac{dx}{x}$

$∫\frac{(logv-1)-1}{v(logv-1)}dv=-∫\frac{dx}{x}$

$∫\frac{dv}{v(logv-1)}-∫\frac{dv}{v(logv-1)}=-∫\frac{dx}{x}$

Let. logv – 1 = z

On differentiating both sides, 

(dv/v) = dz

∫dz – ∫(dz/z) = -∫(dx/x)

z – log|z| = -log|x| + log|c|

(logv – 1) – log|(logv – 1)| = -log|x| + log|c|

logv – log|logv – 1| = -log|x| + log|c| + 1

log|(logv – 1)/v| = log|c₁x|

|logv – 1| = |c₁xv|

|log(y/x) – 1| = |c₁x(y/x)|

|log(y/x) – 1| = |c₁y|  (Where ‘c₁’ is integration constant)

问题36(i)。 (x ² + y ² )dx = 2xydy，y(1)= 0

解决方案：

We have,

 (x² + y²)dx = 2xydy

(dy/dx) = (x² + y²)/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) = (x² + v²x²)/2vx²

v + x(dv/dx) = (1 + v²)/2v

x(dv/dx) = [(1 + v²)/2v] – v

x(dv/dx) = (1 + v² – 2v²)/2v

x(dv/dx) = (1 – v²)/2v

2vdv/(1 – v²) = (dx/x)

On integrating both sides,

∫2vdv/(1 – v²) = ∫(dx/x)

-log|1 – v²| = log|x| – log|c|

log|1 – v²| = log|c/x|

|1 – y²/x²| = |c/x|

|x² – y²| = |cx|

At x = 1, y = 0

1 – 0 = c

c = 1

|x² – y²| = |x|

(x² – y²) = x

问题36(ii)。 xe ^{x / y} – y + x(dy / dx)= 0，y(e)= 0

解决方案：

We have,

xe^x/y – y + x(dy/dx) = 0

(dy/dx) = (y – xe^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 – xe^v)/x

v + x(dv/dx) = v – e^v

x(dv/dx) = v – e^v – v

x(dv/dx) = -e^v

e^-vdv = -(dx/x)

On integrating both sides,

∫e^-vdv = -∫(dx/x)

-e^-v = -log|x| – log|c|

e^-v = log|x| + log|c|

e^-(y/x) = log|x| + log|c|

At x = e, y = 0

e^-(0/e) = log|e| + log|c|

1 = 1 + log|c|

c = 0

e^-y/x = logx

问题36(iii)。 (dy / dx)–(y / x)+ cosec(y / x)= 0，y(1)= 0

解决方案：

We have,

(dy/dx) – (y/x) + cosec(y/x) = 0

(dy/dx) = (y/x) – cosec(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) = v – cosec(v)

x(dv/dx) = v – cosec(v) – v

x(dv/dx) = -cosec(v)

-sin(v)dv = (dx/x)

On integrating both sides,

-∫sin(v)dv = ∫(dx/x)

cos(v) = log|x| + log|c|

cos(y/x) = log|x| + log|c|

At x = 1, y = 0

cos(0/1) = log|1| + log|c|

1 = 0 + log|c|

log|c| = 1

cos(y/x) = log|x| + 1

log|x| = cos(y/x) – 1

问题36(iv)。 (xy – y ² )dx – x ² dy = 0，y(1)= 1

解决方案：

We have,

(xy – y²)dx – x²dy = 0

(dy/dx) = (xy – y²)/x²

(dy/dx) = (y/x) – (y²/x²)

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 – v²

x(dv/dx) = v – v² – v

x(dv/dx) = -v²

-(dv/v²) = (dx/x)

On integrating both sides,

-∫(dv/v²) = ∫(dx/x)

-(-1/v) = log|x| + c

(1/v) = log|x| + c

x/y = log|x| + c

At x = 1, y = 1

1 = log|1| + c

c = 1

x/y = log|x| + 1

y = x/[log|x| + 1]

问题36(v)。 (dy / dx)= [y(x + 2y)] / [x(2x + y)]

解决方案：

We have,

 (dy/dx) = [y(x + 2y)]/[x(2x + 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) = [vx(x + 2vx)]/[x(2x + vx]

x(dv/dx) = [vx(x + 2vx)]/[x(2x + vx] – v

x(dv/dx) = (v + 2v² – 2v – v²)/(2 + v)

x(dv/dx) = (v² – v)/(2 + v)

(2 + v)dv/[v(v – 1)] = (dx/x)

On integrating both sides,

$∫\frac{2+v}{v(v-1)}dv=\frac{dx}{x}$

Using partial derivative,

$\frac{2+v}{v(v-1)}dv=\frac{A}{v}+\frac{B}{v-1}$

2 + v = A(v – 1) + B(v)

2 + v = v(A + B) – A

On  comparing the coefficients,

A = -2

B = 3

-2∫(dv/v) + 3∫dv/(v – 1) = ∫(dx/x)

-2log|v| + 3log|v – 1| = log|x| + log|c|

log|(v – 1)³/v²| = log|xc|

(v – 1)³ = v²|xc|

(y – x)³/x³ = (y/x)²|xc|

(y – x)³ = y²x²c

At x = 1, y = 2,

(2 – 1)³ = 4 * 1 * c

c = (1/4)

(y – x)³ = (1/4)y²x²

问题36(vi)。 (y ⁴ – 2x ³ y)dx +(x ⁴ – 2xy ³ )dy = 0，y(1)= 0

解决方案：

We have,

(y⁴ – 2x³y)dx + (x⁴ – 2xy³)dy = 0

dy/dx = (2x³y – y⁴)/(x⁴ – 2xy³)

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) = (2x³vx – v⁴x⁴)/(x⁴ – 2xv³x³)

v + x(dv/dx) = (2v – v⁴)/(1 – 2v³)

x(dv/dx) = [(2v – v⁴)/(1 – 2v³)] – v

x(dv/dx) = (2v – v⁴ – v + 2v⁴)/(1 – 2v³)

x(dv/dx) = (v + v⁴)/(1 – 2v³)

$\frac{1-2v^3}{v(1+v^3)}dv=\frac{dx}{x}$

$\frac{1+v^3-3v^3}{v(1+v^3)}dv=\frac{dx}{x}$

On integrating both sides,

$∫\frac{1+v^3}{v(1+v^3)}dv-∫\frac{3v^3}{v(1+v^3)}dv=∫\frac{dx}{x}$

∫(dv/v) – ∫(3v²)dv/(1 + v³) = log|x| + log|c|

log|v| – log|1 + v³| = log|xc|

log|v/(1 + v³)| = log|xc|

$\frac{\frac{y}{x}}{1+(\frac{y}{x})^3}=|cx|$

At x = 1, y = 1,

1/(1 + 1) = c

c = (1/2)

(yx²)/(x³ + y³) = (1/2)x

问题36(vii)。 x(x ² + 3y ² )dx + y(y ² + 3x ² )dy = 0，y(1)= 1

解决方案：

We have,

x(x² + 3y²)dx + y(y² + 3x²)dy = 0

dy/dx = -[x(x² + 3y²)/y(y² + 3x²)]

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,

$v+x\frac{dv}{dx}=-\frac{x(x^2+3v^2x^2)}{vx(v^2x^2+3x^2)}$

$v+x\frac{dv}{dx}=-\frac{(1+3v^2)}{v(v^2+3)}$

$x\frac{dv}{dx}=-\frac{(1+3v^2)}{v(v^2+3)}-v$

x(dv/dx) = -(1 + 3v² + v⁴ + 3v²)/v(v² + 3)

[(v³ + 3v)/(1 + 6v² + v⁴)]dv = -(dx/x)

Multiply both sides with 4 and integrating,

$∫\frac{4v^3+12v}{v^4+6v^2+1}dv=-4∫\frac{dx}{x}$

log|v⁴ + 6v² + 1| = -log|x|⁴ + log|c|

|v⁴ + 6v² + 1| = |c/x⁴|

(y⁴ + 6x²y² + x⁴) = c

At y = 1, x = 1

(1 + 6 + 1) = c

c = 8

(y⁴ + 6x²y² + x⁴) = 8

问题36(viii)。 {xsin ² (y / x)– y} dx + xdy = 0，y(1)=π/ 4

解决方案：

We have,

{xsin²(y/x) – y}dx + xdy = 0

dy/dx = [y – xsin²(y/x)]/x

dy/dx = (y/x) – sin²(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) = v – sin²(v)

x(dv/dx) = v – sin²(v) – v

x(dv/dx) = -sin²(v)

-cosec²(v)dv = (dx/x)

On integrating both sides,

-∫cosec²(v) = ∫(dx/x)

cot(v) = log|x| + log|c|

cot(y/x) = log|xc|

At x = 1, y = π/4

cot(π/4) = log|c|

log|c| = 1

c = e

cot(y/x) = log|ex|

问题36(ix)。 x(dy / dx)– y + xsin(y / x)= 0，y(2)=π

解决方案：

We have,

x(dy/dx) – y + xsin(y/x) = 0

x(dy/dx) = y – xsin(y/x)

(dy/dx) = (y/x) – sin(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) = v – sin(v)

x(dv/dx) = v – sin(v) – v

x(dv/dx) = -sin(v)

-cosec(v)dv = (dx/x)

On integrating both sides,

∫cosec(v) = -∫(dx/x)

log|cosec(v) – cot(v)| = -log|x| + log|c|

log|cosec(v) – cot(v)| = -log|x| + log|c|

log|cosec(y/x) – cot(y/x)| = -log|x| + log|c|

At x = 2, y = π

|cosec(π/2) – cot(π/2)| = -log|2| + log|c|

log|c| = 0

log|cosec(y/x) – cot(y/x)| = -log|x|

问题37. xcos(y / x)(dy / dx)= ycos(y / x)+ x，当x = 1时，y =π/ 4

解决方案：

We have,

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

$\frac{dy}{dx}=\frac{ycos(\frac{y}{x})+x}{xcos(\frac{y}{x})}$

(dy/dx) = (y/x) + [1/cos(y/x)]

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 + 1/cosv

x(dv/dx) = v + 1/cosv – v

x(dv/dx) = 1/cosv

cosvdv = (dx/x)

On integrating both sides,

∫cosvdv = ∫(dx/x)

sin(v) = log|x| + log|c|

sin(y/x) = log|x| + log|c|

At x = 1, y = π/4

1/√2 = 0 + log|c|

log|c| = (1/√2)

sin(y/x) = log|x| + (1/√2)

问题38.(x – y)(dy / dx)=(x + 2y)，当x = 1，y = 0时

解决方案：

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(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 + 2vx)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v²)/(1 – v)

(1 – v)dv/(1 + v + v²) = (dx/x)

On integrating both sides,

$∫\frac{1-v}{1+v+v^2}dv=\frac{dx}{x}$

$\frac{1}{2}∫\frac{2-2v}{1+v+v^2}dv=∫\frac{dx}{x}$

$∫\frac{3-(1+2v)}{1+v+v^2}dv=∫\frac{dx}{x}$

$\frac{3}{2}∫\frac{dv}{1+v+v^2}-\frac{1}{2}∫\frac{2v+1}{1+v+v^2}dv=∫\frac{dx}{x}$

$\frac{3}{2}∫\frac{dv}{(v+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}-\frac{1}{2}log|v^2+v+1|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}})-\frac{1}{2}log|v^2+v+1|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|\frac{x^2+xy+y^2}{x^2}|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|x^2+xy+y^2|+log|x|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|x^2+xy+y^2|=c$

At x = 1, y = 0

√3tan^-1|1/√3| – (1/2)log|1| = c

c = √3(π/6)

c = (π/2√3)

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|x^2+xy+y^2|=\frac{π}{2\sqrt{3}}$

$\frac{1}{2}log|x^2+xy+y^2|=2\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{π}{\sqrt{3}}$

问题39.(dy / dx)= xy /(x ² + y ² )

解决方案：

We have,

(dy/dx) = xy/(x² + y²)

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) = xvx/(x² + v²x²)

v + x(dv/dx) = v/(1 + v²)

x(dv/dx) = [v/(1 + v²)] – v

x(dv/dx) = (v – v – v³)/(1 + v²)

[-(1/v³) – (1/v)]dv = (dx/x)

On integrating both sides,

-∫dv/v³ – ∫dv/v = ∫(dx/x)

(1/2v²) – log|v| = log|x| + c

(x²/2y²) = log|vx| + c

(x²/2y²) = log|(y/x)x| + c

(x²/2y²) = log|y| + c

At x = 0, y = 1

c = 0

(x²/2y²) = log|y|