y ′ + x 1 − x 2 y − x y = 0 y'+\dfrac{x}{1-x^2}y-x\sqrt{y}=0 y ′ + 1 − x 2 x y − x y = 0
Let y = u ∗ v y=u*v y = u ∗ v
u ′ v + v ′ u + x 1 − x 2 u v − x u v = 0 u'v+v'u+\dfrac{x}{1-x^2}uv-x\sqrt{uv}=0 u ′ v + v ′ u + 1 − x 2 x uv − x uv = 0
v ( u ′ + x 1 − x 2 u ) + v ′ u − x u v = 0 v(u'+\dfrac{x}{1-x^2}u)+v'u-x\sqrt{uv}=0 v ( u ′ + 1 − x 2 x u ) + v ′ u − x uv = 0
Let first part equal to 0. Then:
{ u ′ + x 1 − x 2 u = 0 v ′ u − x u v = 0 \begin{cases}
u'+\dfrac{x}{1-x^2}u=0\\
v'u-x\sqrt{uv}=0
\end{cases} { u ′ + 1 − x 2 x u = 0 v ′ u − x uv = 0
{ u ′ = − x 1 − x 2 u v ′ u = x u v \begin{cases}
u'=-\dfrac{x}{1-x^2}u\\
v'u=x\sqrt{uv}
\end{cases} { u ′ = − 1 − x 2 x u v ′ u = x uv
{ d u u = − x d x 1 − x 2 d v d x u = x u v \begin{cases}
\dfrac{du}{u}=-\dfrac{xdx}{1-x^2}\\
\dfrac{dv}{dx}u=x\sqrt{uv}
\end{cases} ⎩ ⎨ ⎧ u d u = − 1 − x 2 x d x d x d v u = x uv
{ ln u = ln 1 1 − x 2 d v v = x d x u \begin{cases}
\ln{u}=\ln{\dfrac{1}{\sqrt{1-x^2}}}\\
\dfrac{dv}{\sqrt{v}}=\dfrac{xdx}{\sqrt{u}}
\end{cases} ⎩ ⎨ ⎧ ln u = ln 1 − x 2 1 v d v = u x d x
{ u = 1 1 − x 2 d v v = x 1 − x 2 4 d x \begin{cases}
u=\dfrac{1}{\sqrt{1-x^2}}\\
\dfrac{dv}{\sqrt{v}}=x\sqrt[4]{1-x^2}dx
\end{cases} ⎩ ⎨ ⎧ u = 1 − x 2 1 v d v = x 4 1 − x 2 d x
{ u = 1 1 − x 2 2 v = − 2 5 ( 1 − x 2 ) 5 / 4 + C \begin{cases}
u=\dfrac{1}{\sqrt{1-x^2}}\\
2\sqrt{v}=-\dfrac{2}{5}(1-x^2)^{5/4}+C
\end{cases} ⎩ ⎨ ⎧ u = 1 − x 2 1 2 v = − 5 2 ( 1 − x 2 ) 5/4 + C
{ u = 1 1 − x 2 2 v = − 2 5 ( 1 − x 2 ) 5 / 4 + C \begin{cases}
u=\dfrac{1}{\sqrt{1-x^2}}\\
2\sqrt{v}=-\dfrac{2}{5}(1-x^2)^{5/4}+C
\end{cases} ⎩ ⎨ ⎧ u = 1 − x 2 1 2 v = − 5 2 ( 1 − x 2 ) 5/4 + C
{ u = 1 1 − x 2 v = ( 1 5 ( 1 − x 2 ) 5 / 4 − C 2 ) 2 \begin{cases}
u=\dfrac{1}{\sqrt{1-x^2}}\\
v=(\dfrac{1}{5}(1-x^2)^{5/4}-\dfrac{C}{2})^2
\end{cases} ⎩ ⎨ ⎧ u = 1 − x 2 1 v = ( 5 1 ( 1 − x 2 ) 5/4 − 2 C ) 2
y = 1 1 − x 2 ∗ ( 1 5 ( 1 − x 2 ) 5 / 4 − C 2 ) 2 y=\dfrac{1}{\sqrt{1-x^2}}*(\dfrac{1}{5}(1-x^2)^{5/4}-\dfrac{C}{2})^2 y = 1 − x 2 1 ∗ ( 5 1 ( 1 − x 2 ) 5/4 − 2 C ) 2
y ( 0 ) = 1 = > 1 = 1 1 − 0 2 ∗ ( 1 5 ( 1 − 0 2 ) 5 / 4 − C 2 ) 2 y(0)=1=>\\
1=\dfrac{1}{\sqrt{1-0^2}}*(\dfrac{1}{5}(1-0^2)^{5/4}-\dfrac{C}{2})^2 y ( 0 ) = 1 => 1 = 1 − 0 2 1 ∗ ( 5 1 ( 1 − 0 2 ) 5/4 − 2 C ) 2
1 = 1 1 ∗ ( 1 5 ( 1 ) 5 / 4 − C 2 ) 2 1 = ( 1 5 − C 2 ) 2 = > C = − 8 5 o r С = 12 5 1=\dfrac{1}{\sqrt{1}}*(\dfrac{1}{5}(1)^{5/4}-\dfrac{C}{2})^2\\
1=(\dfrac{1}{5}-\dfrac{C}{2})^2=>\\
C=-\dfrac{8}{5} \space or\space С=\dfrac{12}{5} 1 = 1 1 ∗ ( 5 1 ( 1 ) 5/4 − 2 C ) 2 1 = ( 5 1 − 2 C ) 2 => C = − 5 8 or С = 5 12
y 1 = 1 25 1 − x 2 ∗ ( ( 1 − x 2 ) 5 / 4 + 4 ) 2 y 2 = 1 25 1 − x 2 ∗ ( ( 1 − x 2 ) 5 / 4 − 6 ) 2 y_1=\dfrac{1}{25\sqrt{1-x^2}}*((1-x^2)^{5/4}+4)^2\\
y_2=\dfrac{1}{25\sqrt{1-x^2}}*((1-x^2)^{5/4}-6)^2 y 1 = 25 1 − x 2 1 ∗ (( 1 − x 2 ) 5/4 + 4 ) 2 y 2 = 25 1 − x 2 1 ∗ (( 1 − x 2 ) 5/4 − 6 ) 2
#340153 on Dec 2023
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