Question

find the particular solution of the differential equation,

(refer to attachment)​

Answer 1

Answer:

Given differential equation is

$\sf\dfrac{dy}{dx}=\dfrac{x(2\:log\:x+1)}{sin\:y+y\:cos\:y}$

on seperating the variables, we get

$\sf (sin\:y+y\:cos\:y)dy=x(2\:log\:x+1)dx$

$\sf \implies sin\:y\:dy+y\:cos\:y\:dy=2x\:log\:x\:dx+x\:dx$

on integrating both sides

$\displaystyle\sf \int sin\:y\:dy+\int y\:cos\:y\:dy=2\:\int x \:logx\:dx+\int x\:dx$

$\displaystyle\sf$

$\displaystyle\sf \implies -cos\:y+\left[y\int cos\:y\:dy-\int \left\{ \dfrac{d}{dy}(y)\int cos\:y\:dy \right\} dy \right]$

$\displaystyle\sf$

$\displaystyle\sf = 2\left[ logx\int x \ dx - \int \left\{ \dfrac{d}{dx}logx \int x \ dx\right\}dx\right]+\dfrac{x^2}{2}$

$\displaystyle\sf$

$\displaystyle\sf\implies -cos\:y+y\:sin\:y-\int sin\:y\:dy = 2\left[ \dfrac{x^2}{2} logx - \int \left\{ \dfrac{1}{x}\dfrac{x^2}{2}\right\}dx\right] + \dfrac{x^2}{2}$

$\displaystyle\sf$

$\displaystyle\sf\implies -cos\:y+y\:sin\:y+cos\:y=x^2\:logx-\int x\:dx + \dfrac{x^2}{2}$

$\displaystyle\sf$

$\displaystyle\sf\implies y\:sin\:y = x^2\;logx-\dfrac{x^2}{2}+\dfrac{x^2}{2}+C$

$\displaystyle\sf$

$\displaystyle\sf \implies y\:sin\:y = x^2 \: logx + C \:\;\;\;\dots (i)$

$\displaystyle\sf$

$\displaystyle\sf\pink\bigstar$Also, given that y = π/2 when x = 1.

on putting these values

$\displaystyle\sf$

$\displaystyle\sf\dfrac{\pi}{2}\:sin\left(\dfrac{\pi}{2}\right) = 1^2\:log(1)+C$

$\displaystyle\sf \implies C = \dfrac{\pi}{2}$

on substituting the value of C in 1, we get

$\boxed{\displaystyle\sf{\bf{y\:sin\:y= x^2\:logx+\dfrac{\boldsymbol\pi}{2}}}}$

which is the required solution.

@Delìnqúentìnvíncìblé

_________________________________

Answer 2

Step-by-step explanation:

Given differential equation is

\sf\dfrac{dy}{dx}=\dfrac{x(2\:log\:x+1)}{sin\:y+y\:cos\:y}

dx

dy

=

siny+ycosy

x(2logx+1)

on seperating the variables, we get

\sf (sin\:y+y\:cos\:y)dy=x(2\:log\:x+1)dx(siny+ycosy)dy=x(2logx+1)dx

\sf \implies sin\:y\:dy+y\:cos\:y\:dy=2x\:log\:x\:dx+x\:dx⟹sinydy+ycosydy=2xlogxdx+xdx

on integrating both sides

\displaystyle\sf \int sin\:y\:dy+\int y\:cos\:y\:dy=2\:\int x \:logx\:dx+\int x\:dx∫sinydy+∫ycosydy=2∫xlogxdx+∫xdx

\displaystyle\sf

\displaystyle\sf \implies -cos\:y+\left[y\int cos\:y\:dy-\int \left\{ \dfrac{d}{dy}(y)\int cos\:y\:dy \right\} dy \right]⟹−cosy+[y∫cosydy−∫{

dy

d

(y)∫cosydy}dy]

\displaystyle\sf

\displaystyle\sf = 2\left[ logx\int x \ dx - \int \left\{ \dfrac{d}{dx}logx \int x \ dx\right\}dx\right]+\dfrac{x^2}{2}=2[logx∫x dx−∫{

dx

d

logx∫x dx}dx]+

2

x

2

\displaystyle\sf

\displaystyle\sf\implies -cos\:y+y\:sin\:y-\int sin\:y\:dy = 2\left[ \dfrac{x^2}{2} logx - \int \left\{ \dfrac{1}{x}\dfrac{x^2}{2}\right\}dx\right] + \dfrac{x^2}{2}⟹−cosy+ysiny−∫sinydy=2[

2

x

2

logx−∫{

x

1

2

x

2

}dx]+

2

x

2

\displaystyle\sf

\displaystyle\sf\implies -cos\:y+y\:sin\:y+cos\:y=x^2\:logx-\int x\:dx + \dfrac{x^2}{2}⟹−cosy+ysiny+cosy=x

2

logx−∫xdx+

2

x

2

\displaystyle\sf

\displaystyle\sf\implies y\:sin\:y = x^2\;logx-\dfrac{x^2}{2}+\dfrac{x^2}{2}+C⟹ysiny=x

2

logx−

2

x

2

+

2

x

2

+C

\displaystyle\sf

\displaystyle\sf \implies y\:sin\:y = x^2 \: logx + C \:\;\;\;\dots (i)⟹ysiny=x

2

logx+C…(i)

\displaystyle\sf

\displaystyle\sf\pink\bigstar★ Also, given that y = π/2 when x = 1.

on putting these values

\displaystyle\sf

\displaystyle\sf\dfrac{\pi}{2}\:sin\left(\dfrac{\pi}{2}\right) = 1^2\:log(1)+C

2

π

sin(

2

π

)=1

2

log(1)+C

\displaystyle\sf \implies C = \dfrac{\pi}{2}⟹C=

2

π

on substituting the value of C in 1, we get

\boxed{\displaystyle\sf{\bf{y\:sin\:y= x^2\:logx+\dfrac{\boldsymbol\pi}{2}}}}

ysiny=x

2

logx+

2

π

which is the required solution.

@Delìnqúentìnvíncìblé