Answer:-
1.
y ( 3 x 3 − x + y ) d x + x 2 ( 1 − x 2 ) d y = 0 y(3x^3-x+y)dx+x^2(1-x^2)dy=0 y ( 3 x 3 − x + y ) d x + x 2 ( 1 − x 2 ) d y = 0 This is the first order nonlinear ordinary differential equation.
Divide on x 2 ( 1 − x 2 ) d x . x^2(1-x^2)dx. x 2 ( 1 − x 2 ) d x .
y ′ = 3 x 3 y − x y + y 2 x 2 ( x 2 − 1 ) , y'=\frac{3x^3y-xy+y^2}{x^2(x^2-1)}, y ′ = x 2 ( x 2 − 1 ) 3 x 3 y − x y + y 2 , y ′ = y 2 x 2 ( x − 1 ) ( x + 1 ) + 3 x y ( x − 1 ) ( x + 1 ) − y x ( x − 1 ) ( x + 1 ) , y'=\frac{y^2}{x^2(x-1)(x+1)}+\frac{3xy}{(x-1)(x+1)}-\frac{y}{x(x-1)(x+1)}, y ′ = x 2 ( x − 1 ) ( x + 1 ) y 2 + ( x − 1 ) ( x + 1 ) 3 x y − x ( x − 1 ) ( x + 1 ) y , y ′ − 3 x 2 − 1 x 3 − x y = y 2 x 4 − x 2 . y'-\frac{3x^2-1}{x^3-x}y=\frac{y^2}{x^4-x^2}. y ′ − x 3 − x 3 x 2 − 1 y = x 4 − x 2 y 2 . We have Bernuli's equation for n=2.
Divide by y 2 . y^2. y 2 .
y ′ y 2 − 3 x 2 − 1 ( x 3 − x ) y = 1 x 4 − x 2 . \frac{y'}{y^2}-\frac{3x^2-1}{(x^3-x)y}=\frac{1}{x^4-x^2}. y 2 y ′ − ( x 3 − x ) y 3 x 2 − 1 = x 4 − x 2 1 . Replacement u = 1 y , u=\frac{1}{y}, u = y 1 , u ′ = − y ′ y 2 , y ′ = − u ′ y 2 . u'=-\frac{y'}{y^2}, y'=-u'y^2. u ′ = − y 2 y ′ , y ′ = − u ′ y 2 .
− u ( 3 x 2 − 1 ) x 3 − x − u ′ = 1 x 4 − x 2 , -\frac{u(3x^2-1)}{x^3-x}-u'=\frac{1}{x^4-x^2}, − x 3 − x u ( 3 x 2 − 1 ) − u ′ = x 4 − x 2 1 , u ′ + ( 3 x ( x − 1 ) ( x + 1 ) − 1 x ( x − 1 ) ( x + 1 ) ) u = − 1 x 4 − x 2 u'+(\frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)})u=-\frac{1}{x^4-x^2} u ′ + ( ( x − 1 ) ( x + 1 ) 3 x − x ( x − 1 ) ( x + 1 ) 1 ) u = − x 4 − x 2 1 This is the first order linear equation.
Let be P ( x ) = 3 x ( x − 1 ) ( x + 1 ) − 1 x ( x − 1 ) ( x + 1 ) . P(x)=\frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)}. P ( x ) = ( x − 1 ) ( x + 1 ) 3 x − x ( x − 1 ) ( x + 1 ) 1 .
∫ P ( x ) d x = ∫ 3 x ( x − 1 ) ( x + 1 ) − 1 x ( x − 1 ) ( x + 1 ) d x = ln ( x 3 − x ) + C , \int P(x)dx=\int \frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)}dx=\ln (x^3-x)+C, ∫ P ( x ) d x = ∫ ( x − 1 ) ( x + 1 ) 3 x − x ( x − 1 ) ( x + 1 ) 1 d x = ln ( x 3 − x ) + C ,
where C is some constant.
Solve
u ′ + P ( x ) u = 0 , u'+P(x)u=0, u ′ + P ( x ) u = 0 , d u u = − P ( x ) d x , \frac{du}{u}=-P(x)dx, u d u = − P ( x ) d x , ∫ d u u = − ∫ P ( x ) d x , \int\frac{du}{u}=-\int P(x)dx, ∫ u d u = − ∫ P ( x ) d x , ln u = − ln ( x 3 − x ) + C , \ln u =-\ln (x^3-x)+C, ln u = − ln ( x 3 − x ) + C , u = C x 3 − x . u=\frac{C}{x^3-x}. u = x 3 − x C . The solution of u ′ + ( 3 x ( x − 1 ) ( x + 1 ) − 1 x ( x − 1 ) ( x + 1 ) ) u = − 1 x 4 − x 2 u'+(\frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)})u=-\frac{1}{x^4-x^2} u ′ + ( ( x − 1 ) ( x + 1 ) 3 x − x ( x − 1 ) ( x + 1 ) 1 ) u = − x 4 − x 2 1 u = C ( x ) x 3 − x . u=\frac{C(x)}{x^3-x}. u = x 3 − x C ( x ) .
C ( x ) = ∫ ( − x 3 − x x 4 − x 2 ) d x = − ln x + C . C(x)=\int(-\frac{x^3-x}{x^4-x^2})dx=-\ln x+C. C ( x ) = ∫ ( − x 4 − x 2 x 3 − x ) d x = − ln x + C . Thus u = − ln x + C x 3 − x . u=\frac{-\ln x+C}{x^3-x}. u = x 3 − x − l n x + C .
From here y = x − x 3 ln x + C . y=\frac{x-x^3}{\ln x+C}. y = l n x + C x − x 3 .
Answer. y = x − x 3 ln x + C . y=\frac{x-x^3}{\ln x+C}. y = l n x + C x − x 3 .
2.
x ( y 2 − x ) d y = y ( y 2 + 2 x ) d x = 0. x(y^2-x)dy=y(y^2+2x)dx=0. x ( y 2 − x ) d y = y ( y 2 + 2 x ) d x = 0. This is the first order nonlinear ordinary differential equation.
Replacement y = z , y=\sqrt{z}, y = z , d y = d z 2 z . dy=\frac{dz}{2\sqrt{z}}. d y = 2 z d z .
( c z 2 − x 3 2 z ) d z = ( − z 3 2 − 2 x z ) d x . (\frac{c\sqrt{z}}{2}-\frac{x^3}{2\sqrt z})dz=(-z^{\frac{3}{2}}-2x\sqrt z)dx. ( 2 c z − 2 z x 3 ) d z = ( − z 2 3 − 2 x z ) d x . Replacement u = z x , u=\frac{z}{x}, u = x z ,
x = u x , d z = u d x + x d u . x=ux, dz=udx+xdu. x = ux , d z = u d x + x d u .
We will have
( u 2 − 1 2 u x 3 / 2 ( u d x + x d u ) = ( − u 3 / 2 − 2 u ) x 3 / 2 d x , (\frac{\sqrt {u}}{2}-\frac{1}{2\sqrt u}x^{3/2}(udx+xdu)=(-u^{3/2}-2\sqrt u)x^{3/2}dx, ( 2 u − 2 u 1 x 3/2 ( u d x + x d u ) = ( − u 3/2 − 2 u ) x 3/2 d x , ( u x 5 / 2 2 − x 5 / 2 2 u ) d u = ( − 3 u 3 / 2 x 3 / 2 2 − 3 u u 3 / 2 2 ) d x . (\frac{\sqrt {u}x^{5/2}}{2}-\frac{x^{5/2}}{2\sqrt u})du=(-\frac{3u^{3/2}x^{3/2}}{2}-\frac{3\sqrt u u^{3/2}}{2})dx. ( 2 u x 5/2 − 2 u x 5/2 ) d u = ( − 2 3 u 3/2 x 3/2 − 2 3 u u 3/2 ) d x . Divide by x 5 / 2 x^{5/2} x 5/2 − 3 u 3 / 2 x 3 / 2 2 − 3 u u 3 / 2 2 . -\frac{3u^{3/2}x^{3/2}}{2}-\frac{3\sqrt u u^{3/2}}{2}. − 2 3 u 3/2 x 3/2 − 2 3 u u 3/2 .
We will have
( 1 3 u ( u + 1 ) − 1 3 ( u + 1 ) ) d u = d x x . (\frac{1}{3u(u+1)}-\frac{1}{3(u+1)})du=\frac{dx}{x}. ( 3 u ( u + 1 ) 1 − 3 ( u + 1 ) 1 ) d u = x d x . ∫ ( 1 3 u ( u + 1 ) − 1 3 ( u + 1 ) ) d u = ∫ d x x . \int(\frac{1}{3u(u+1)}-\frac{1}{3(u+1)})du=\int\frac{dx}{x}. ∫ ( 3 u ( u + 1 ) 1 − 3 ( u + 1 ) 1 ) d u = ∫ x d x .
ln u 3 − 2 ln ( u + 1 ) 3 = ln x + C , \frac{\ln u}{3}-\frac{2\ln (u+1)}{3}=\ln x+C, 3 l n u − 3 2 l n ( u + 1 ) = ln x + C ,
where C is some constant.
From here
u ( u + 1 ) 2 = C x 3 . \frac{u}{(u+1)^2}=Cx^3. ( u + 1 ) 2 u = C x 3 . Thus
z ( z + x ) 2 = C x 2 . \frac{z}{(z+x)^2}=Cx^2. ( z + x ) 2 z = C x 2 . So we have answer
y 2 = C x 2 ( y 2 + x ) 2 . y^2=Cx^2(y^2+x)^2. y 2 = C x 2 ( y 2 + x ) 2 . Answer. y 2 = C x 2 ( y 2 + x ) 2 . y^2=Cx^2(y^2+x)^2. y 2 = C x 2 ( y 2 + x ) 2 .
#340153 on Dec 2023
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