Answer on Question #58884 - Math - Differential Equations
Question
8. Solve
y ( x y + 1 ) d x + x ( 1 + x y + x 2 y 2 ) d y = 0 y(xy + 1)dx + x(1 + xy + x^2y^2)dy = 0 y ( x y + 1 ) d x + x ( 1 + x y + x 2 y 2 ) d y = 0 Solution
Transform the equation:
y ′ = − y ( x y + 1 ) x ( 1 + x y + x 2 y 2 ) y' = -\frac{y(xy + 1)}{x(1 + xy + x^2y^2)} y ′ = − x ( 1 + x y + x 2 y 2 ) y ( x y + 1 )
Substitute:
u ( x ) = x y ⇒ y = u x ⇒ y ′ = u ′ x − u x 2 . u(x) = xy \quad \Rightarrow \quad y = \frac{u}{x} \quad \Rightarrow \quad y' = \frac{u'x - u}{x^2}. u ( x ) = x y ⇒ y = x u ⇒ y ′ = x 2 u ′ x − u .
Next,
u ′ x − u x 2 + u ( u + 1 ) x 2 ( u 2 + u + 1 ) = 0 , \frac{u'x - u}{x^2} + \frac{u(u + 1)}{x^2(u^2 + u + 1)} = 0, x 2 u ′ x − u + x 2 ( u 2 + u + 1 ) u ( u + 1 ) = 0 , ( u ′ x − u ) ( u 2 + u + 1 ) + u ( u + 1 ) = 0 , (u'x - u)(u^2 + u + 1) + u(u + 1) = 0, ( u ′ x − u ) ( u 2 + u + 1 ) + u ( u + 1 ) = 0 , u ′ x ( u 2 + u + 1 ) = u ( u 2 + u + 1 ) − u ( u + 1 ) , u'x(u^2 + u + 1) = u(u^2 + u + 1) - u(u + 1), u ′ x ( u 2 + u + 1 ) = u ( u 2 + u + 1 ) − u ( u + 1 ) , u ′ x ( u 2 + u + 1 ) = u 3 , u'x(u^2 + u + 1) = u^3, u ′ x ( u 2 + u + 1 ) = u 3 , ( 1 u + 1 u 2 + 1 u 3 ) d u = d x x , \left(\frac{1}{u} + \frac{1}{u^2} + \frac{1}{u^3}\right)du = \frac{dx}{x}, ( u 1 + u 2 1 + u 3 1 ) d u = x d x , ln u − 1 u − 1 2 u 2 = ln x + C . \ln u - \frac{1}{u} - \frac{1}{2u^2} = \ln x + C. ln u − u 1 − 2 u 2 1 = ln x + C .
Substituting u = x y u = xy u = x y
ln ( x y ) − 1 x y − 1 2 x 2 y 2 = ln x + C . \ln(xy) - \frac{1}{xy} - \frac{1}{2x^2y^2} = \ln x + C. ln ( x y ) − x y 1 − 2 x 2 y 2 1 = ln x + C .
Answer: ln ( x y ) − 1 x y − 1 2 x 2 y 2 = ln x + C . \ln(xy) - \frac{1}{xy} - \frac{1}{2x^2y^2} = \ln x + C. ln ( x y ) − x y 1 − 2 x 2 y 2 1 = ln x + C .
Question
9. Solve
x d y − y d x − x 2 − y 2 d x = 0 xdy - ydx - \sqrt{x^2 - y^2}dx = 0 x d y − y d x − x 2 − y 2 d x = 0
Solution
x y ′ = y + x 2 − y 2 , y = t x ⇒ y ′ = t ′ x + t , x ( t ′ x + t ) = t x + x 1 − t 2 , t ′ x = 1 − t 2 , d t 1 − t 2 = d x x , sin − 1 t = ln ∣ x ∣ + ln ∣ C ∣ . \begin{array}{l}
x y ^ {\prime} = y + \sqrt {x ^ {2} - y ^ {2}}, \\
y = t x \quad \Rightarrow \quad y ^ {\prime} = t ^ {\prime} x + t, \\
x (t ^ {\prime} x + t) = t x + x \sqrt {1 - t ^ {2}}, \\
t ^ {\prime} x = \sqrt {1 - t ^ {2}}, \\
\frac {d t}{\sqrt {1 - t ^ {2}}} = \frac {d x}{x}, \\
\sin^ {- 1} t = \ln | x | + \ln | C |.
\end{array} x y ′ = y + x 2 − y 2 , y = t x ⇒ y ′ = t ′ x + t , x ( t ′ x + t ) = t x + x 1 − t 2 , t ′ x = 1 − t 2 , 1 − t 2 d t = x d x , sin − 1 t = ln ∣ x ∣ + ln ∣ C ∣.
Using the previous equality and formula t = y x t = \frac{y}{x} t = x y
sin − 1 y x = ln ∣ C x ∣ . \sin^ {- 1} \frac {y}{x} = \ln | C x |. sin − 1 x y = ln ∣ C x ∣.
Answer:
sin − 1 y x = ln ∣ C x ∣ . \sin^ {- 1} \frac {y}{x} = \ln | C x |. sin − 1 x y = ln ∣ C x ∣.
Question
10. The population of student p p p
Solution
The rate of increasing population:
p ′ = d p d t = k ( p + 150250 + p 3 ) . p ^ {\prime} = \frac {d p}{d t} = k \left(p + 1 5 0 2 5 0 + \frac {p}{3}\right). p ′ = d t d p = k ( p + 150250 + 3 p ) .
The differential equation is
p ′ = k ( 4 p 3 + 150250 ) . p ^ {\prime} = k \left(\frac {4 p}{3} + 1 5 0 2 5 0\right). p ′ = k ( 3 4 p + 150250 ) .
Answer:
p ′ = k ( 4 p 3 + 150250 ) . p ^ {\prime} = k \left(\frac {4 p}{3} + 1 5 0 2 5 0\right). p ′ = k ( 3 4 p + 150250 ) .
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