y ′ + f ( x ) y = g ( x ) y α / : y α y ′ y α + f ( x ) y y α = g ( x ) y ′ y α + f ( x ) y 1 − α = g ( x ) смена : y 1 − α = z ( 1 − α ) y 1 − α − 1 y ′ = z ′ ( 1 − α ) y ′ y α = z ′ y ′ y α = z ′ 1 − α z ′ 1 − α + f ( x ) z = g ( x ) / ⋅ ( 1 − α ) z ′ + ( 1 − α ) f ( x ) z = ( 1 − α ) g ( x ) \begin{array}{l} y' + f\left( x \right)y = g\left( x \right){y^\alpha }{\rm{ }} {\text{ }}{\text{ }} /:{y^\alpha } \\ \frac{{y'}}{{{y^\alpha }}} + f\left( x \right)\frac{y}{{{y^\alpha }}} = g\left( x \right) \\ \frac{{y'}}{{{y^\alpha }}} + f\left( x \right){y^{1 - \alpha }} = g\left( x \right) \\ смена:{y^{1 - \alpha }} = z \\ {\text{ }} {\text{ }}{\text{ }}{\text{ }}\left( {1 - \alpha } \right){y^{1 - \alpha - 1}}y' = z' \\ {\text{ }}{\text{ }}{\text{ }}{\text{ }} \left( {1 - \alpha } \right)\frac{{y'}}{{{y^\alpha }}} = z' \\ {\text{ }}{\text{ }}{\text{ }}{\text{ }} \frac{{y'}}{{{y^\alpha }}} = \frac{{z'}}{{1 - \alpha }} \\ \\ \frac{{z'}}{{1 - \alpha }} + f\left( x \right)z = g\left( x \right){\rm{ }}{\text{ }}{\text{ }} / \cdot \left( {1 - \alpha } \right) \\ z' + \left( {1 - \alpha } \right)f\left( x \right)z = \left({1 - \alpha } \right)g\left( x \right) \end{array} y ′ + f ( x ) y = g ( x ) y α / : y α y α y ′ + f ( x ) y α y = g ( x ) y α y ′ + f ( x ) y 1 − α = g ( x ) смена : y 1 − α = z ( 1 − α ) y 1 − α − 1 y ′ = z ′ ( 1 − α ) y α y ′ = z ′ y α y ′ = 1 − α z ′ 1 − α z ′ + f ( x ) z = g ( x ) / ⋅ ( 1 − α ) z ′ + ( 1 − α ) f ( x ) z = ( 1 − α ) g ( x )
Задаци 1. Одреди опште решење диференцијалне једначине y ′ + 2 x y = 2 x 3 y 3 y' + 2xy = 2{x^3}{y^3} y ′ + 2 x y = 2 x 3 y 3
Ово је Бернулијева диференцијална једначина, задата у њеном стандардном облику, код које је α = 3 \alpha = 3 α = 3 Дакле, најпре диференцијалну једначину делимо са y 3 {y^3} y 3 y ′ + 2 x y = 2 x 3 y 3 / : y 3 y' + 2xy = 2{x^3}{y^3}{\rm{ }}{\text{ }}{\text{ }} /:{y^3} y ′ + 2 x y = 2 x 3 y 3 / : y 3
y ′ y 3 + 2 x 1 y 2 = 2 x 3 \frac{{y'}}{{{y^3}}} + 2x\frac{1}{{{y^2}}} = 2{x^3} y 3 y ′ + 2 x y 2 1 = 2 x 3
Уводимо смену 1 y 2 = z \frac{1}{{{y^2}}} = z y 2 1 = z y − 2 = z {y^{ - 2}} = z y − 2 = z Како је y y y − 2 y − 3 y ′ = z ′ - 2{y^{ - 3}}y' = z' − 2 y − 3 y ′ = z ′ y ′ y 3 = z ′ − 2 \frac{{y'}}{{{y^3}}} = \frac{{z'}}{{ - 2}} y 3 y ′ = − 2 z ′
Уврштавајући то у полазну диференцијалну једначину z ′ − 2 + 2 x z = 2 x 3 / ⋅ ( − 2 ) \frac{{z'}}{{ - 2}} + 2xz = 2{x^3}{\rm{ }} {\text{ }}{\text{ }} / \cdot \left( { - 2} \right) − 2 z ′ + 2 x z = 2 x 3 / ⋅ ( − 2 ) z ′ − 4 x z = − 4 x 3 z' - 4xz = - 4{x^3} z ′ − 4 x z = − 4 x 3 z = u v ⇒ z ′ = u ′ v + u v ′ z = uv{\text{ }}{\text{ }} \Rightarrow{\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ − 4 x u v = − 4 x 3 u'v + uv' - 4xuv = - 4{x^3} u ′ v + u v ′ − 4 xuv = − 4 x 3
u ′ v + u ( v ′ − 4 x v ) = − 4 x 3 u'v + u\left( {v' - 4xv} \right) = - 4{x^3} u ′ v + u ( v ′ − 4 xv ) = − 4 x 3
v ′ − 4 x v = 0 v' - 4xv = 0 v ′ − 4 xv = 0
d v d x = 4 x v \frac{{dv}}{{dx}} = 4xv d x d v = 4 xv
d v v = 4 x d x / ∫ \frac{{dv}}{v} = 4xdx{\rm{ }}{\text{ }}{\text{ }} /\int {} v d v = 4 x d x / ∫
ln ∣ v ∣ = 4 x 2 2 \ln \left| v \right| = 4\frac{{{x^2}}}{2} ln ∣ v ∣ = 4 2 x 2
v = e 2 x 2 v = {e^{2{x^2}}} v = e 2 x 2
u ′ e 2 x 2 = − 4 x 3 u'{e^{2{x^2}}} = - 4{x^3} u ′ e 2 x 2 = − 4 x 3
d u d x = − 4 x 3 e − 2 x 2 \frac{{du}}{{dx}} = - 4{x^3}{e^{ - 2{x^2}}} d x d u = − 4 x 3 e − 2 x 2
d u = − 4 x 3 e − 2 x 2 d x / ∫ du = - 4{x^3}{e^{ - 2{x^2}}}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} d u = − 4 x 3 e − 2 x 2 d x / ∫
u = − 4 ∫ x 3 e − 2 x 2 d x = ∣ u = x 2 d v = x e − 2 x 2 d x d u = 2 x d x v = ∫ x e − 2 x 2 d x = ∣ − 2 x 2 = t − 4 x d x = d t ∣ v = ∫ e t d t − 4 = − 1 4 e − 2 x 2 ∣ = u = - 4\int {{x^3}{e^{ - 2{x^2}}}dx} = \left| {\begin{array}{cccccccccccccccccccc} {u = {x^2}}&{dv = x{e^{ - 2{x^2}}}dx}\\ {du = 2xdx}&{v = \int {x{e^{ - 2{x^2}}}dx = \left| {\begin{array}{cccccccccccccccccccc} { - 2{x^2} = t}\\ { - 4xdx = dt} \end{array}} \right|} }\\ {}&{v = \int {{e^t}\frac{{dt}}{{ - 4}} = - \frac{1}{4}{e^{ - 2{x^2}}}} } \end{array}} \right| = u = − 4 ∫ x 3 e − 2 x 2 d x = u = x 2 d u = 2 x d x d v = x e − 2 x 2 d x v = ∫ x e − 2 x 2 d x = − 2 x 2 = t − 4 x d x = d t v = ∫ e t − 4 d t = − 4 1 e − 2 x 2 =
= − 4 ( − x 2 4 e − 2 x 2 + 2 4 ∫ x e − 2 x 2 d x ) {\text{ }}{\text{ }} = - 4\left( { - \frac{{{x^2}}}{4}{e^{ - 2{x^2}}} + \frac{2}{4}\int {x{e^{ - 2{x^2}}}dx} } \right) = − 4 ( − 4 x 2 e − 2 x 2 + 4 2 ∫ x e − 2 x 2 d x )
= x 2 e − 2 x 2 − 2 ( − 1 4 e − 2 x 2 ) + C = x 2 e − 2 x 2 + 1 2 e − 2 x 2 + C {\text{ }}{\text{ }} = {x^2}{e^{ - 2{x^2}}} - 2\left( { - \frac{1}{4}{e^{ - 2{x^2}}}} \right) + C = {x^2}{e^{ - 2{x^2}}} + \frac{1}{2}{e^{ - 2{x^2}}} + C = x 2 e − 2 x 2 − 2 ( − 4 1 e − 2 x 2 ) + C = x 2 e − 2 x 2 + 2 1 e − 2 x 2 + C
z = u v = x 2 + 1 2 + C e 2 x 2 z = uv = {x^2} + \frac{1}{2} + C{e^{2{x^2}}} z = uv = x 2 + 2 1 + C e 2 x 2
Како је 1 y 2 = z ⇒ y = 1 z \frac{1}{{{y^2}}} = z{\rm{ }} \Rightarrow {\rm{ }}y = \frac{1}{{\sqrt z }} y 2 1 = z ⇒ y = z 1 y = 1 x 2 + 1 2 + C e 2 x 2 y = \frac{1}{{\sqrt {{x^2} + \frac{1}{2} + C{e^{2{x^2}}}} }} y = x 2 + 2 1 + C e 2 x 2 1
2. Наћи оно партикуларно решење диференцијалне једначине x y ′ + y = − x y 2 xy' + y = - x{y^2} x y ′ + y = − x y 2 y ( 1 ) = 1 2 y\left( 1 \right) = \frac{1}{2} y ( 1 ) = 2 1
Најпре једначину делимо са x x x x y ′ + y = − x y 2 / : x xy' + y = - x{y^2} {\text{ }}{\text{ }} {\rm{ /:}}x x y ′ + y = − x y 2 / : x
y ′ + 1 x y = − y 2 / : y 2 y' + \frac{1}{x}y = - {y^2}{\text{ }}{\text{ }} {\rm{ /:}}{y^2} y ′ + x 1 y = − y 2 / : y 2
y ′ y 2 + 1 x 1 y = − 1 \frac{{y'}}{{{y^2}}} + \frac{1}{x}\frac{1}{y} = - 1 y 2 y ′ + x 1 y 1 = − 1
1 y = z ⇒ y − 1 = z ⇒ − y − 2 y ′ = z ′ ⇒ y ′ y 2 = − z ′ \frac{1}{y} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} {y^{ - 1}} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} - {y^{ - 2}}y' = z'{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} \frac{{y'}}{{{y^2}}} = - z' y 1 = z ⇒ y − 1 = z ⇒ − y − 2 y ′ = z ′ ⇒ y 2 y ′ = − z ′
− z ′ + 1 x z = − 1 / : ( − 1 ) - z' + \frac{1}{x}z = - 1{\text{ }}{\text{ }} {\rm{ /:}}\left( { - 1} \right) − z ′ + x 1 z = − 1 / : ( − 1 )
z ′ − 1 x z = 1 z' - \frac{1}{x}z = 1 z ′ − x 1 z = 1
z = u v ⇒ z ′ = u ′ v + u v ′ z = uv {\text{ }}{\text{ }} \Rightarrow {\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ − 1 x u v = 1 u'v + uv' - \frac{1}{x}uv = 1 u ′ v + u v ′ − x 1 uv = 1
u ′ v + u ( v ′ − 1 x v ) = 1 u'v + u\left( {v' - \frac{1}{x}v} \right) = 1 u ′ v + u ( v ′ − x 1 v ) = 1
v ′ − 1 x v = 0 v' - \frac{1}{x}v = 0 v ′ − x 1 v = 0
d v d x = 1 x v \frac{{dv}}{{dx}} = \frac{1}{x}v d x d v = x 1 v
d v v = 1 x d x / ∫ \frac{{dv}}{v} = \frac{1}{x}dx {\text{ }}{\text{ }} {\rm{ /}}\int {} v d v = x 1 d x / ∫
ln ∣ v ∣ = ln ∣ x ∣ \ln \left| v \right| = \ln \left| x \right| ln ∣ v ∣ = ln ∣ x ∣
v = x v = x v = x
u ′ x = 1 u'x = 1 u ′ x = 1
d u d x = 1 x \frac{{du}}{{dx}} = \frac{1}{x} d x d u = x 1
d u = 1 x d x / ∫ du = \frac{1}{x}dx{\text{ }}{\text{ }} {\rm{ /}}\int {} d u = x 1 d x / ∫
u = ln ∣ x ∣ + ln C u = \ln \left| x \right| + \ln C u = ln ∣ x ∣ + ln C
u = ln x C u = \ln xC u = ln x C
z = u v = x ln x C z = uv = x\ln xC z = uv = x ln x C
y = 1 z = 1 x ln x C y = \frac{1}{z} = \frac{1}{{x\ln xC}} y = z 1 = x ln x C 1
Уврштавајући почетни услов 1 2 = 1 1 ⋅ ln 1 ⋅ C ⇒ ln C = 2 ⇒ C = e 2 \frac{1}{2} = \frac{1}{{1 \cdot \ln 1 \cdot C}} \Rightarrow \ln C = 2 \Rightarrow C = {e^2} 2 1 = 1 ⋅ l n 1 ⋅ C 1 ⇒ ln C = 2 ⇒ C = e 2
y = 1 x ( ln x + 2 ) y = \frac{1}{{x\left( {\ln x + 2} \right)}} y = x ( ln x + 2 ) 1
3. Одреди опште решење диференцијалне једначине y ′ + 2 x y = y 3 x 2 y' + \frac{2}{x}y = \frac{{{y^3}}}{{{x^2}}} y ′ + x 2 y = x 2 y 3
y ′ + 2 x y = y 3 x 2 / : y 3 y' + \frac{2}{x}y = \frac{{{y^3}}}{{{x^2}}}{\rm{ }}{\text{ }}{\text{ }} /:{y^3} y ′ + x 2 y = x 2 y 3 / : y 3
y ′ y 3 + 2 x 1 y 2 = 1 x 2 \frac{{y'}}{{{y^3}}} + \frac{2}{x}\frac{1}{{{y^2}}} = \frac{1}{{{x^2}}} y 3 y ′ + x 2 y 2 1 = x 2 1
1 y 2 = z ⇒ y − 2 = z ⇒ − 2 y − 3 y ′ = z ′ ⇒ y ′ y 3 = z ′ − 2 \frac{1}{{{y^2}}} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} {y^{ - 2}} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} - 2{y^{ - 3}}y' = z'{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} \frac{{y'}}{{{y^3}}} = \frac{{z'}}{{ - 2}} y 2 1 = z ⇒ y − 2 = z ⇒ − 2 y − 3 y ′ = z ′ ⇒ y 3 y ′ = − 2 z ′
z ′ − 2 + 2 x z = 1 x 2 / ⋅ ( − 2 ) \frac{{z'}}{{ - 2}} + \frac{2}{x}z = \frac{1}{{{x^2}}}{\rm{ }}{\text{ }}{\text{ }} / \cdot \left( { - 2} \right) − 2 z ′ + x 2 z = x 2 1 / ⋅ ( − 2 )
z ′ − 4 x z = − 2 x 2 z' - \frac{4}{x}z = - \frac{2}{{{x^2}}} z ′ − x 4 z = − x 2 2
z = u v ⇒ z ′ = u ′ v + u v ′ z = uv{\text{ }}{\text{ }} \Rightarrow {\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ − 4 x u v = − 2 x 2 u'v + uv' - \frac{4}{x}uv = - \frac{2}{{{x^2}}} u ′ v + u v ′ − x 4 uv = − x 2 2
u ′ v + u ( v ′ − 4 x v ) = − 2 x 2 u'v + u\left( {v' - \frac{4}{x}v} \right) = - \frac{2}{{{x^2}}} u ′ v + u ( v ′ − x 4 v ) = − x 2 2
v ′ − 4 x v = 0 v' - \frac{4}{x}v = 0 v ′ − x 4 v = 0
d v d x = 4 x v \frac{{dv}}{{dx}} = \frac{4}{x}v d x d v = x 4 v
d v v = 4 x d x / ∫ \frac{{dv}}{v} = \frac{4}{x}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} v d v = x 4 d x / ∫
ln ∣ v ∣ = 4 ln ∣ x ∣ \ln \left| v \right| = 4\ln \left| x \right| ln ∣ v ∣ = 4 ln ∣ x ∣
v = x 4 v = {x^4} v = x 4
u ′ x 4 = − 2 x 2 u'{x^4} = - \frac{2}{{{x^2}}} u ′ x 4 = − x 2 2
d u d x = − 2 x 6 \frac{{du}}{{dx}} = - \frac{2}{{{x^6}}} d x d u = − x 6 2
d u = − 2 x 6 d x / ∫ du = - \frac{2}{{{x^6}}}dx{\rm{ }} {\text{ }}{\text{ }} /\int {} d u = − x 6 2 d x / ∫
u = − 2 ∫ x − 6 d x u = - 2\int {{x^{ - 6}}dx} u = − 2 ∫ x − 6 d x
u = 2 5 x 5 + C u = \frac{2}{{5{x^5}}} + C u = 5 x 5 2 + C
z = u v = 2 5 x + C x 4 z = uv = \frac{2}{{5x}} + C{x^4} z = uv = 5 x 2 + C x 4
y = 1 z = 1 2 5 x + C x 4 y = \frac{1}{{\sqrt z }} = \frac{1}{{\sqrt {\frac{2}{{5x}} + C{x^4}} }} y = z 1 = 5 x 2 + C x 4 1
4. Одреди опште решење диференцијалне једначине x y ′ + y = y 2 ln x xy' + y = {y^2}\ln x x y ′ + y = y 2 ln x
x y ′ + y = y 2 ln x / : x xy' + y = {y^2}\ln x{\rm{ }}{\text{ }}{\text{ }} /:x x y ′ + y = y 2 ln x / : x
y ′ + 1 x y = y 2 ln x x / : y 2 y' + \frac{1}{x}y = {y^2}\frac{{\ln x}}{x}{\rm{ }}{\text{ }}{\text{ }} /:{y^2} y ′ + x 1 y = y 2 x l n x / : y 2
y ′ y 2 + 1 x 1 y = ln x x \frac{{y'}}{{{y^2}}} + \frac{1}{x}\frac{1}{y} = \frac{{\ln x}}{x} y 2 y ′ + x 1 y 1 = x l n x
1 y = z ⇒ y − 1 = z ⇒ − y − 2 y ′ = z ′ ⇒ y ′ y 2 = − z ′ \frac{1}{y} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} {y^{ - 1}} = z{\rm{ }} {\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} - {y^{ - 2}}y' = z'{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} \frac{{y'}}{{{y^2}}} = - z' y 1 = z ⇒ y − 1 = z ⇒ − y − 2 y ′ = z ′ ⇒ y 2 y ′ = − z ′
− z ′ + 1 x z = ln x x / ⋅ ( − 1 ) - z' + \frac{1}{x}z = \frac{{\ln x}}{x}{\rm{ }}{\text{ }}{\text{ }} / \cdot \left( { - 1} \right) − z ′ + x 1 z = x l n x / ⋅ ( − 1 )
z ′ − 1 x z = − ln x x z' - \frac{1}{x}z = - \frac{{\ln x}}{x} z ′ − x 1 z = − x l n x
z = u v ⇒ z ′ = u ′ v + u v ′ z = uv{\text{ }}{\text{ }} \Rightarrow {\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ − 1 x u v = − ln x x u'v + uv' - \frac{1}{x}uv = - \frac{{\ln x}}{x} u ′ v + u v ′ − x 1 uv = − x l n x
u ′ v + u ( v ′ − 1 x v ) = − ln x x u'v + u\left( {v' - \frac{1}{x}v} \right) = - \frac{{\ln x}}{x} u ′ v + u ( v ′ − x 1 v ) = − x l n x
v ′ − 1 x v = 0 v' - \frac{1}{x}v = 0 v ′ − x 1 v = 0
d v d x = 1 x v \frac{{dv}}{{dx}} = \frac{1}{x}v d x d v = x 1 v
d v v = 1 x d x / ∫ \frac{{dv}}{v} = \frac{1}{x}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} v d v = x 1 d x / ∫
ln ∣ v ∣ = ln ∣ x ∣ \ln \left| v \right| = \ln \left| x \right| ln ∣ v ∣ = ln ∣ x ∣
v = x v = x v = x
u ′ x = − ln x x u'x = - \frac{{\ln x}}{x} u ′ x = − x l n x
d u d x = − ln x x 2 \frac{{du}}{{dx}} = - \frac{{\ln x}}{{{x^2}}} d x d u = − x 2 l n x
d u = − ln x x 2 d x / ∫ du = - \frac{{\ln x}}{{{x^2}}}dx{\rm{ }} {\text{ }}{\text{ }} /\int {} d u = − x 2 l n x d x / ∫
u = − ∫ ln x x 2 d x = ∣ ln x = t ⇒ x = e t 1 x d x = d t ∣ = − ∫ t e t d t = u = - \int {\frac{{\ln x}}{{{x^2}}}dx} = \left| {\begin{array}{cccccccccccccccccccc} {\ln x = t \Rightarrow x = {e^t}}\\ {\frac{1}{x}dx = dt} \end{array}} \right| = - \int {\frac{t}{{{e^t}}}} dt = u = − ∫ x 2 l n x d x = ln x = t ⇒ x = e t x 1 d x = d t = − ∫ e t t d t =
= − ∫ t e − t d t = ∣ u = t d v = e − t d t d u = d t v = − e − t ∣ = {\text{ }}{\text{ }} = - \int {t{e^{ - t}}dt} = \left| {\begin{array}{cccccccccccccccccccc} {u = t}&{dv = {e^{ - t}}dt}\\ {du = dt}&{v = - {e^{ - t}}} \end{array}} \right| = = − ∫ t e − t d t = u = t d u = d t d v = e − t d t v = − e − t =
= − ( − t e t + ∫ e − t d t ) = t e t + e t + C = {\text{ }}{\text{ }} = - \left( { - t{e^t} + \int {{e^{ - t}}dt} } \right) = t{e^t} + {e^t} + C = = − ( − t e t + ∫ e − t d t ) = t e t + e t + C =
= x ln x + x + C {\text{ }}{\text{ }} = x\ln x + x + C = x ln x + x + C
z = u v = x 2 ln x + x 2 + C x z = uv = {x^2}\ln x + {x^2} + Cx z = uv = x 2 ln x + x 2 + C x
y = 1 z = 1 x 2 ln x + x 2 + C x y = \frac{1}{z} = \frac{1}{{{x^2}\ln x + {x^2} + Cx}} y = z 1 = x 2 ln x + x 2 + C x 1
5. Одреди опште решење диференцијалне једначине x y ′ − 4 y − x 2 y xy' - 4y - {x^2}\sqrt y x y ′ − 4 y − x 2 y
x y ′ − 4 y − x 2 y = 0 / : x xy' - 4y - {x^2}\sqrt y = 0{\rm{ }} {\text{ }}{\text{ }} /:x x y ′ − 4 y − x 2 y = 0 / : x
y ′ − 4 x y = x y / : y y' - \frac{4}{x}y = x\sqrt y {\rm{ }}{\text{ }}{\text{ }} /:\sqrt y y ′ − x 4 y = x y / : y
y ′ y − 4 x y = x \frac{{y'}}{{\sqrt y }} - \frac{4}{x}\sqrt y = x y y ′ − x 4 y = x
y = z ⇒ y 1 2 = z ⇒ 1 2 y − 1 2 y ′ = z ′ ⇒ y ′ y = 2 z ′ \sqrt y = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} {y^{\frac{1}{2}}} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} \frac{1}{2}{y^{ - \frac{1}{2}}}y' = z'{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} \frac{{y'}}{{\sqrt y }} = 2z' y = z ⇒ y 2 1 = z ⇒ 2 1 y − 2 1 y ′ = z ′ ⇒ y y ′ = 2 z ′
2 z ′ − 4 x z = x / : 2 2z' - \frac{4}{x}z = x{\rm{ }}{\text{ }}{\text{ }} /:2 2 z ′ − x 4 z = x / : 2
z ′ − 2 x z = x 2 z' - \frac{2}{x}z = \frac{x}{2} z ′ − x 2 z = 2 x
z = u v ⇒ z ′ = u ′ v + u v ′ z = uv {\text{ }}{\text{ }} \Rightarrow {\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ − 2 x u v = x 2 u'v + uv' - \frac{2}{x}uv = \frac{x}{2} u ′ v + u v ′ − x 2 uv = 2 x
u ′ v + u ( v ′ − 2 x v ) = x 2 u'v + u\left( {v' - \frac{2}{x}v} \right) = \frac{x}{2} u ′ v + u ( v ′ − x 2 v ) = 2 x
v ′ − 2 x v = 0 v' - \frac{2}{x}v = 0 v ′ − x 2 v = 0
d v d x = 2 x v \frac{{dv}}{{dx}} = \frac{2}{x}v d x d v = x 2 v
d v v = 2 x d x / ∫ \frac{{dv}}{v} = \frac{2}{x}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} v d v = x 2 d x / ∫
ln ∣ v ∣ = 2 ln ∣ x ∣ \ln \left| v \right| = 2\ln \left| x \right| ln ∣ v ∣ = 2 ln ∣ x ∣
v = x 2 v = {x^2} v = x 2
u ′ x 2 = x 2 u'{x^2} = \frac{x}{2} u ′ x 2 = 2 x
d u d x = 1 2 x \frac{{du}}{{dx}} = \frac{1}{{2x}} d x d u = 2 x 1
d u = 1 2 x d x / ∫ du = \frac{1}{{2x}}dx{\rm{ }} {\text{ }}{\text{ }} /\int {} d u = 2 x 1 d x / ∫
u = 1 2 ln ∣ x ∣ + 1 2 ln C u = \frac{1}{2}\ln \left| x \right| + \frac{1}{2}\ln C u = 2 1 ln ∣ x ∣ + 2 1 ln C
u = ln x C u = \ln \sqrt {xC} u = ln x C
z = u v = x 2 ln x C z = uv = {x^2}\ln \sqrt {xC} z = uv = x 2 ln x C
y = z 2 = x 4 ln 2 x C y = {z^2} = {x^4}{\ln ^2}\sqrt {xC} y = z 2 = x 4 ln 2 x C
6. Одреди оно партикуларно решење диференцијалне једначине x d x = ( x 2 y − y 3 ) d y xdx = \left( {\frac{{{x^2}}}{y} - {y^3}} \right)dy x d x = ( y x 2 − y 3 ) d y y ( 2 ) = 1 y\left( 2 \right) = 1 y ( 2 ) = 1
Ову диференцијалну једначину није могуће свести на Бернулијеву диференцијалну једначину по y y y x x x x ′ + f ( y ) x = g ( y ) x α x' + f\left( y \right)x = g\left( y \right){x^\alpha } x ′ + f ( y ) x = g ( y ) x α x d x d y = x 2 y − y 3 x\frac{{dx}}{{dy}} = \frac{{{x^2}}}{y} - {y^3} x d y d x = y x 2 − y 3
x x ′ = x 2 y − y 3 / : x xx' = \frac{{{x^2}}}{y} - {y^3}{\rm{ }}{\text{ }}{\text{ }} /:x x x ′ = y x 2 − y 3 / : x
x ′ = 1 y x − y 3 x x' = \frac{1}{y}x - \frac{{{y^3}}}{x} x ′ = y 1 x − x y 3
x ′ − 1 y x = − y 3 x x' - \frac{1}{y}x = - \frac{{{y^3}}}{x} x ′ − y 1 x = − x y 3
x ′ − 1 y x = − y 3 x / ⋅ x x' - \frac{1}{y}x = - \frac{{{y^3}}}{x}{\rm{ }}{\text{ }}{\text{ }} / \cdot x x ′ − y 1 x = − x y 3 / ⋅ x
x x ′ − 1 y x 2 = − y 3 xx' - \frac{1}{y}{x^2} = - {y^3} x x ′ − y 1 x 2 = − y 3
x 2 = z ⇒ 2 x x ′ = z ′ ⇒ x x ′ = z ′ 2 {x^2} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\text{ }}{\text{ }} {\rm{ 2}}xx' = z'{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} xx' = \frac{{z'}}{2} x 2 = z ⇒ 2 x x ′ = z ′ ⇒ x x ′ = 2 z ′
z ′ 2 − 1 y z = − y 3 / ⋅ 2 \frac{{z'}}{2} - \frac{1}{y}z = - {y^3}{\rm{ }}{\text{ }}{\text{ }} / \cdot 2 2 z ′ − y 1 z = − y 3 / ⋅ 2
z ′ − 2 y z = − 2 y 3 z' - \frac{2}{y}z = - 2{y^3} z ′ − y 2 z = − 2 y 3
z = u v ⇒ z ′ = u ′ v + u v ′ z = uv{\text{ }}{\text{ }} \Rightarrow{\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ − 2 y u v = − 2 y 3 u'v + uv' - \frac{2}{y}uv = - 2{y^3} u ′ v + u v ′ − y 2 uv = − 2 y 3
u ′ v + u ( v ′ − 2 y v ) = − 2 y 3 u'v + u\left( {v' - \frac{2}{y}v} \right) = - 2{y^3} u ′ v + u ( v ′ − y 2 v ) = − 2 y 3
v ′ − 2 y v = 0 v' - \frac{2}{y}v = 0 v ′ − y 2 v = 0
d v d y = 2 y v \frac{{dv}}{{dy}} = \frac{2}{y}v d y d v = y 2 v
d v v = 2 y d y / ∫ \frac{{dv}}{v} = \frac{2}{y}dy{\rm{ }} {\text{ }}{\text{ }} /\int {} v d v = y 2 d y / ∫
ln ∣ v ∣ = 2 ln ∣ y ∣ \ln \left| v \right| = 2\ln \left| y \right| ln ∣ v ∣ = 2 ln ∣ y ∣
v = y 2 v = {y^2} v = y 2
u ′ y 2 = − 2 y 3 u'{y^2} = - 2{y^3} u ′ y 2 = − 2 y 3
d u d y = − 2 y \frac{{du}}{{dy}} = - 2y d y d u = − 2 y
d u = − 2 y d y / ∫ du = - 2ydy{\rm{ }} {\text{ }}{\text{ }} /\int {} d u = − 2 y d y / ∫
u = − 2 y 2 2 + C u = - 2\frac{{{y^2}}}{2} + C u = − 2 2 y 2 + C
u = − y 2 + C u = - {y^2} + C u = − y 2 + C
z = − y 4 + C y 2 z = - {y^4} + C{y^2} z = − y 4 + C y 2
x = z = − y 4 + C y 2 = y C − y 2 x = \sqrt z = \sqrt { - {y^4} + C{y^2}} = y\sqrt {C - {y^2}} x = z = − y 4 + C y 2 = y C − y 2
Уврштавајући почетни услов 2 = 1 ⋅ C − 1 ⇒ C = 5 2 = 1 \cdot \sqrt {C - 1} \Rightarrow C = 5 2 = 1 ⋅ C − 1 ⇒ C = 5
x = y 5 − y 2 x = y\sqrt {5 - {y^2}} x = y 5 − y 2
7. Одреди опште решење диференцијалне једначине y ′ − y t g x + y 2 cos x = 0 y' - ytgx + {y^2}\cos x = 0 y ′ − y t g x + y 2 cos x = 0
y ′ − y t g x = − y 2 cos x / : y 2 y' - ytgx = - {y^2}\cos x{\rm{ }}{\text{ }}{\text{ }}/:{y^2} y ′ − y t g x = − y 2 cos x / : y 2
y ′ y 2 − t g x ⋅ 1 y = − cos x \frac{{y'}}{{{y^2}}} - tgx \cdot \frac{1}{y} = - \cos x y 2 y ′ − t g x ⋅ y 1 = − cos x
1 y = z ⇒ y − 1 = z ⇒ − y − 2 y ′ = z ′ ⇒ y ′ y 2 = − z ′ \frac{1}{y} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} {y^{ - 1}} = z{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} - {y^{ - 2}}y' = z'{\rm{ }}{\text{ }}{\text{ }} \Rightarrow {\rm{ }}{\text{ }}{\text{ }} \frac{{y'}}{{{y^2}}} = - z' y 1 = z ⇒ y − 1 = z ⇒ − y − 2 y ′ = z ′ ⇒ y 2 y ′ = − z ′
− z ′ − t g x ⋅ z = − cos x / ⋅ ( − 1 ) - z' - tgx \cdot z = - \cos x{\rm{ }}{\text{ }}{\text{ }}/ \cdot \left( { - 1} \right) − z ′ − t g x ⋅ z = − cos x / ⋅ ( − 1 )
z ′ + t g x ⋅ z = cos x z' + tgx \cdot z = \cos x z ′ + t g x ⋅ z = cos x
z = u v ⇒ z ′ = u ′ v + u v ′ z = uv{\text{ }}{\text{ }} \Rightarrow {\text{ }}{\text{ }} z' = u'v + uv' z = uv ⇒ z ′ = u ′ v + u v ′
u ′ v + u v ′ + t g x u v = cos x u'v + uv' + tgxuv = \cos x u ′ v + u v ′ + t g xuv = cos x
u ′ v + u ( v ′ + t g x v ) = cos x u'v + u\left( {v' + tgxv} \right) = \cos x u ′ v + u ( v ′ + t g xv ) = cos x
v ′ + t g x v = 0 v' + tgxv = 0 v ′ + t g xv = 0
d v d x = − t g x v \frac{{dv}}{{dx}} = - tgxv d x d v = − t g xv
d v v = − t g x d x / ∫ \frac{{dv}}{v} = - tgxdx{\rm{ }}{\text{ }}{\text{ }}/\int {} v d v = − t g x d x / ∫
∫ d v v = − ∫ sin x cos x d x \int {\frac{{dv}}{v}} = - \int {\frac{{\sin x}}{{\cos x}}dx} ∫ v d v = − ∫ c o s x s i n x d x
∣ cos x = t − sin x d x = d t ∣ {\text{ }}{\text{ }}{\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} \left| {\begin{array}{cccccccccccccccccccc} {\cos x = t}\\ { - \sin xdx = dt} \end{array}} \right| cos x = t − sin x d x = d t
ln ∣ v ∣ = ∫ d t t \ln \left| v \right| = \int {\frac{{dt}}{t}} ln ∣ v ∣ = ∫ t d t
ln ∣ v ∣ = ln ∣ t ∣ \ln \left| v \right| = \ln \left| t \right| ln ∣ v ∣ = ln ∣ t ∣
v = t v = t v = t
v = cos x v = \cos x v = cos x
u ′ cos x = cos x u'\cos x = \cos x u ′ cos x = cos x
d u d x = 1 \frac{{du}}{{dx}} = 1 d x d u = 1
d u = d x / ∫ du = dx{\rm{ }}{\text{ }}{\text{ }}/\int {} d u = d x / ∫
u = x + C u = x + C u = x + C
z = u v = ( x + C ) cos x z = uv = \left( {x + C} \right)\cos x z = uv = ( x + C ) cos x
y = 1 z = 1 ( x + C ) cos x y = \frac{1}{z} = \frac{1}{{\left( {x + C} \right)\cos x}} y = z 1 = ( x + C ) cos x 1