Question 1: x d y d x − ( x + 1 ) y = x y 2 x\frac{dy}{dx}-(x+1)y=xy^2 x d x d y − ( x + 1 ) y = x y 2
Solution: Dividing equation by − x y 2 -xy^2 − x y 2 :
− y − 2 y ′ + x + 1 x y − 1 = − 1 -y^{-2}y'+\frac{x+1}xy^{-1}=-1 − y − 2 y ′ + x x + 1 y − 1 = − 1
Substituting z = y 1 − 2 = y − 1 z=y^{1-2}=y^{-1} z = y 1 − 2 = y − 1 , d z d x = z ′ = − y − 2 y ′ \frac{dz}{dx}=z'=-y^{-2}y' d x d z = z ′ = − y − 2 y ′ , therefore: (1)
z ′ + x + 1 x z = − 1 z'+\frac{x+1}{x}z=-1 z ′ + x x + 1 z = − 1
Substituting z = u v z=uv z = uv , z ′ = u ′ v + u v ′ z'=u'v+uv' z ′ = u ′ v + u v ′ : (2)
u ′ v + u v ′ + x + 1 x u v = − 1 u'v+uv'+\frac{x+1}{x}uv=-1 u ′ v + u v ′ + x x + 1 uv = − 1
u ′ v + u ( v ′ + x + 1 x v ) = − 1 u'v+u(v'+\frac{x+1}{x}v)=-1 u ′ v + u ( v ′ + x x + 1 v ) = − 1 (3)
Setting the part inside () to 0 and separating variables:
v ′ + x + 1 x v = 0 v'+\frac{x+1}{x}v=0 v ′ + x x + 1 v = 0
d v d x = − x + 1 x v \frac{dv}{dx}=- \frac{x+1}{x}v d x d v = − x x + 1 v
∫ d v v = − ∫ x + 1 x d x \int\frac{dv}v = -\int\frac{x+1}{x}dx ∫ v d v = − ∫ x x + 1 d x
l n ( v ) = − ( x + l n ( x ) + l n ( k ) ) ln(v)=-(x+ln(x)+ln(k)) l n ( v ) = − ( x + l n ( x ) + l n ( k ))
v = e − ( x + l n ( x ) ) = 1 k x e x v=e^{-(x+ln(x))}=\frac1 {kxe^x} v = e − ( x + l n ( x )) = k x e x 1 (4)
Putting v v v from (4) into equation (3):
1 k x e x u ′ = − 1 \frac1 {kxe^x}u'=-1 k x e x 1 u ′ = − 1
∫ d u = − k ∫ x e x \int du=-k\int xe^x ∫ d u = − k ∫ x e x
u = − k ( x e x − e x + e x + C ) = − k ( e x ( x − 1 ) + C ) u=-k(xe^x-e^x+e^x+C)=-k(e^x(x-1)+C) u = − k ( x e x − e x + e x + C ) = − k ( e x ( x − 1 ) + C )
Finding z from (2):
z = u v = − ( e x ( x − 1 ) + C ) x e x z=uv=\frac{-(e^x(x-1)+C)}{xe^x} z = uv = x e x − ( e x ( x − 1 ) + C )
Finding y from (1):
y = z − 1 = − x e x e x ( x − 1 ) + C y=z^{-1}=-\frac{xe^x}{e^x(x-1)+C} y = z − 1 = − e x ( x − 1 ) + C x e x
Answer 1: y = − x e x e x ( x − 1 ) + C y=-\frac{xe^x}{e^x(x-1)+C} y = − e x ( x − 1 ) + C x e x
Question 2: x 2 y ′ − 2 x y = 4 y 4 x^2y'-2xy=4y^4 x 2 y ′ − 2 x y = 4 y 4 , y ( 1 ) = 1 2 y(1)=\frac 12 y ( 1 ) = 2 1
Solution: Dividing equation by − 1 3 x 2 y 4 -\frac1 3 x^2 y^4 − 3 1 x 2 y 4 :
− 3 y − 4 y ′ + 6 x y − 3 = − 12 x 2 -3y^{-4}y'+\frac 6 x y^{-3}=-\frac {12} {x^2} − 3 y − 4 y ′ + x 6 y − 3 = − x 2 12
Substituting z = y − 3 z=y^{-3} z = y − 3 , z ′ = − 3 y − 4 y ′ z'=-3y^{-4}y' z ′ = − 3 y − 4 y ′ , therefore: (1)
z ′ + 6 x z = − 12 x 2 z'+\frac6xz=-\frac{12}{x^2} z ′ + x 6 z = − x 2 12
Substituting z = u v , z ′ = u ′ v + u v ′ z=uv , z'=u'v+uv' z = uv , z ′ = u ′ v + u v ′ : (2)
u ′ v + u v ′ + 6 x u v = − 12 x 2 u'v+uv'+\frac6xuv=-\frac{12}{x^2} u ′ v + u v ′ + x 6 uv = − x 2 12
u ′ v + u ( v ′ + 6 x v ) = − 12 x 2 u'v+u(v'+\frac6xv)=-\frac{12}{x^2} u ′ v + u ( v ′ + x 6 v ) = − x 2 12 (3)
Setting the part inside () to 0 and separating variables:
v ′ + 6 x v = 0 v'+\frac6xv=0 v ′ + x 6 v = 0 (4)
v ′ = − 6 x v v'=-\frac6xv v ′ = − x 6 v
∫ d v v = − 6 ∫ d x x \int \frac {dv} v = -6 \int \frac {dx} x ∫ v d v = − 6 ∫ x d x
l n ( v ) = − 6 l n ( x ) − l n ( k ) = l n ( k x − 6 ) ln(v)= -6 ln(x)-ln(k)=ln(kx^{-6}) l n ( v ) = − 6 l n ( x ) − l n ( k ) = l n ( k x − 6 )
v = 1 k x 6 v=\frac 1 {kx^6} v = k x 6 1
Putting v v v from (4) into equation (3):
1 k x 6 u ′ = − 12 x 2 \frac 1 {kx^6}u'=-\frac{12}{x^2} k x 6 1 u ′ = − x 2 12
∫ d u = ∫ − 12 k x 4 d x \int du = \int -12kx^4dx ∫ d u = ∫ − 12 k x 4 d x
u = k ( − 12 x 5 5 + C ) u= k(-\frac {12x^5}5+C) u = k ( − 5 12 x 5 + C )
Finding z from (2):
z = u v = − ( 12 x 5 5 + C ) ∗ 1 x 6 z=uv=-(\frac {12x^5}5+C)*\frac1{x^6} z = uv = − ( 5 12 x 5 + C ) ∗ x 6 1
Finding y from (1):
y = 5 x 6 12 x 5 + 5 C 3 y=\sqrt[3]{\frac{5x^6}{12x^5+5C}} y = 3 12 x 5 + 5 C 5 x 6
Assuming y ( 1 ) = 1 2 y(1)=\frac 12 y ( 1 ) = 2 1 : 5 ∗ ( 1 ) 6 12 ∗ ( 1 ) 5 + 5 C 3 = 1 2 \sqrt[3]{\frac{5*(1)^6}{12*(1)^5+5C}}=\frac12 3 12 ∗ ( 1 ) 5 + 5 C 5 ∗ ( 1 ) 6 = 2 1
5 12 + 5 C 3 = 1 2 \sqrt[3]{\frac{5}{12+5C}}=\frac12 3 12 + 5 C 5 = 2 1
5 12 + 5 C = 1 8 \frac{5}{12+5C}=\frac18 12 + 5 C 5 = 8 1
C = 52 5 C=\frac{52}5 C = 5 52
Substitute C into general solution:
y = 5 x 6 12 x 5 + 52 3 y=\sqrt[3]{\frac{5x^6}{12x^5+52}} y = 3 12 x 5 + 52 5 x 6
Answer 2: y = 5 x 6 12 x 5 + 52 3 y=\sqrt[3]{\frac{5x^6}{12x^5+52}} y = 3 12 x 5 + 52 5 x 6
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