Q u e s t i o n : f i n d t h e i n t e g r a l s u r f a c e o f t h e e q u a t i o n ( y − z ) 2 x y p + ( x 2 − y 2 ) q + z ( x 2 − y 2 ) = 0 t h r o u g h t h e c u r v e x = t 2 , y = 0 , z = t 3 s o l u t i o n : u s e l a g r a n g e ′ s m e t h o d d x ( y − z ) 2 x y = d y ( y − z ) ( x 2 − y 2 ) = d z y 2 − x 2 N o w I = I I d x ( y − z ) 2 x y = d y ( y − z ) ( x 2 − y 2 ) ( x 2 − y 2 ) d x − 2 x y d y = 0 t h i s i s e x a c t e q u a t i o n ( s i n c e i f M d x + N d y = 0 , a n d ∂ M ∂ y = ∂ N ∂ x t h e n i t i s c a l l e d e x a c t e q u a t i o n . N o w w e k n o w t h a t f o r e x a c t e q u a t i o n s , s o l u t i o n i s ∫ M d x + ∫ [ N − ∂ ∫ M d x ∂ y ] d y = c 1 h e n c e x 3 3 − y 2 x = c 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) N o w I I = I I I d y = ( z − y ) d z d y d z + y = z i t i s l i n e a r d i f f . e q . / / f o r i t ′ s s o l u t i o n I . F . = e ∫ P d z , w h e r e p = 1 m u l t i p l e o f y I . F . = e ∫ 1 d z = e z a n d s o l u t i o n a s e z . y = ∫ e z . z d z e z . y = e z ( z − 1 ) + c 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 ) n o w w e h a v e 2 e q u a t i o n s x 3 3 − y 2 x = c 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) e z . y = e z ( z − 1 ) + c 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 ) g i v e n c u r v e x = t 2 , y = 0 , z = t 3 ; p u t t h i s v a l u e s i n e q u a t i o n s ; ( t 2 ) 3 3 − ( 0 ) 2 ( t 2 ) = c 1 c 1 = ( t 2 ) 3 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 4 ) a n d e z . y = e z ( z − 1 ) + c 2 e t 3 . ( 0 ) = e t 3 ( t 3 − 1 ) + c 2 c 2 = e t 3 ( 1 − t 3 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 5 ) n o w p u t v a l u e o f t 3 = 3 c 1 i n ( 5 ) c 2 = e 3 c 1 ( 1 − 3 c 1 ) n o w p u t v a l u e o f c 1 a n d c 2 e z . y − e z ( z − 1 ) = e 3 ( x 3 3 − y 2 x ) ( 1 − 3 ( x 3 3 − y 2 x ) ) Question:\\ find \space the\space integral\space surface \space of \space the\space equation \\(y-z){2xyp+(x^2-y^2)q}+z(x^2-y^2)=0\\
through\space the\space curve\\ x=t^2, y=0 , z=t^3\\
solution:
\\
use\space lagrange's \space method
\\
\frac{dx}{(y-z)2xy}=\frac{dy}{(y-z)(x^2-y^2)}=\frac{dz}{y^2-x^2}\\
Now \space I=II\\
\frac{dx}{(y-z)2xy}=\frac{dy}{(y-z)(x^2-y^2)}\\
{(x^2-y^2)}{dx}-{2xy}{dy}=0\\
this \space is\space exact\space equation\space\\
(since\space if\space Mdx+Ndy=0,\space and\space \frac{∂M}{∂y}=\frac{∂N}{∂x}\space then\space it \space is \space called\space exact \space equation.
Now\space we \space know\space that\\
for\space exact \space equations\space ,solution \space is\space\\
\int Mdx+\int [N-\frac{∂\int Mdx}{∂y}] dy=c_1\\
hence\\
\frac{x^3}{3}-y^2x=c_1.................................(1)\\
Now \space II=III\\
dy=(z-y)dz\\
\frac{dy}{dz}+y=z\\
it \space is \space linear \space diff.\space eq.//
for \space it's \space solution\space \\
I.F.=e^{\int P dz}, \space where \space p=1\space multiple\space of \space y\\
I.F.=e^{\int 1 dz}={e}^{z}\\
and \\
solution\space as\\
{e}^{z}.y=\int {e}^{z}.z \space dz
{e}^{z} .y= {e}^{z}(z-1)+c_2..............................(2)\\
now\space we \space have\space 2\space equations\\
\frac{x^3}{3}-y^2x=c_1.................................(1)\\
{e}^{z} .y= {e}^{z}(z-1)+c_2..............................(2)\\
given \space curve
x=t^2, y=0 , z=t^3;
put \space this\space values\space in\space equations;
\frac{(t^2)^3}{3}-(0)^2(t^2)=c_1\\
c_1=\frac{(t^2)^3}{3}................................(4)\\
and\\
{e}^{z} .y= {e}^{z}(z-1)+c_2\\
{e}^{t^3} .(0)= {e}^{t^3}(t^3-1)+c_2\\
c_2={e}^{t^3}(1-t^3)...........................(5)\\
now\space put\space value\space of \space t^3=\sqrt{3c_1}\space in\space (5)\\
c_2={e}^{\sqrt{3c_1}}(1-\sqrt{3c_1})\\
now\space put\space value\space of \space c_1\space and\space c_2\\
{e}^{z} .y- {e}^{z}(z-1)={e}^{\sqrt{3(\frac{x^3}{3}-y^2x)}}(1-\sqrt{3(\frac{x^3}{3}-y^2x)})\\ Q u es t i o n : f in d t h e in t e g r a l s u r f a ce o f t h e e q u a t i o n ( y − z ) 2 x y p + ( x 2 − y 2 ) q + z ( x 2 − y 2 ) = 0 t h ro ug h t h e c u r v e x = t 2 , y = 0 , z = t 3 so l u t i o n : u se l a g r an g e ′ s m e t h o d ( y − z ) 2 x y d x = ( y − z ) ( x 2 − y 2 ) d y = y 2 − x 2 d z N o w I = II ( y − z ) 2 x y d x = ( y − z ) ( x 2 − y 2 ) d y ( x 2 − y 2 ) d x − 2 x y d y = 0 t hi s i s e x a c t e q u a t i o n ( s in ce i f M d x + N d y = 0 , an d ∂ y ∂ M = ∂ x ∂ N t h e n i t i s c a ll e d e x a c t e q u a t i o n . N o w w e kn o w t ha t f or e x a c t e q u a t i o n s , so l u t i o n i s ∫ M d x + ∫ [ N − ∂ y ∂ ∫ M d x ] d y = c 1 h e n ce 3 x 3 − y 2 x = c 1 ................................. ( 1 ) N o w II = III d y = ( z − y ) d z d z d y + y = z i t i s l in e a r d i ff . e q .// f or i t ′ s so l u t i o n I . F . = e ∫ P d z , w h ere p = 1 m u lt i pl e o f y I . F . = e ∫ 1 d z = e z an d so l u t i o n a s e z . y = ∫ e z . z d z e z . y = e z ( z − 1 ) + c 2 .............................. ( 2 ) n o w w e ha v e 2 e q u a t i o n s 3 x 3 − y 2 x = c 1 ................................. ( 1 ) e z . y = e z ( z − 1 ) + c 2 .............................. ( 2 ) g i v e n c u r v e x = t 2 , y = 0 , z = t 3 ; p u t t hi s v a l u es in e q u a t i o n s ; 3 ( t 2 ) 3 − ( 0 ) 2 ( t 2 ) = c 1 c 1 = 3 ( t 2 ) 3 ................................ ( 4 ) an d e z . y = e z ( z − 1 ) + c 2 e t 3 . ( 0 ) = e t 3 ( t 3 − 1 ) + c 2 c 2 = e t 3 ( 1 − t 3 ) ........................... ( 5 ) n o w p u t v a l u e o f t 3 = 3 c 1 in ( 5 ) c 2 = e 3 c 1 ( 1 − 3 c 1 ) n o w p u t v a l u e o f c 1 an d c 2 e z . y − e z ( z − 1 ) = e 3 ( 3 x 3 − y 2 x ) ( 1 − 3 ( 3 x 3 − y 2 x ) )
#340153 on Dec 2023
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