Answer on Question #84948 – Math – Differential Equations
Question
Solve the following differential equation
x p 2 − 2 y p + a x = 0 x p ^ {2} - 2 y p + a x = 0 x p 2 − 2 y p + a x = 0
Solution
x p 2 − 2 y p + a x = 0 x p ^ {2} - 2 y p + a x = 0 x p 2 − 2 y p + a x = 0 2 y p = x p 2 + a x 2 y p = x p ^ {2} + a x 2 y p = x p 2 + a x y = x p 2 + a x 2 p y = \frac {x p ^ {2} + a x}{2 p} y = 2 p x p 2 + a x y = x p 2 + a x 2 p y = \frac {x p}{2} + \frac {a x}{2 p} y = 2 x p + 2 p a x
Differentiate both sides with respect to x x x
d y d x = 1 2 ( p + x d p d x ) + a 2 ( 1 p − x p 2 d p d x ) \frac {d y}{d x} = \frac {1}{2} \left(p + x \frac {d p}{d x}\right) + \frac {a}{2} \left(\frac {1}{p} - \frac {x}{p ^ {2}} \frac {d p}{d x}\right) d x d y = 2 1 ( p + x d x d p ) + 2 a ( p 1 − p 2 x d x d p )
Then
p = p 2 + x 2 ⋅ d p d x + a 2 p − a x 2 p 2 ⋅ d p d x p = \frac {p}{2} + \frac {x}{2} \cdot \frac {d p}{d x} + \frac {a}{2 p} - \frac {a x}{2 p ^ {2}} \cdot \frac {d p}{d x} p = 2 p + 2 x ⋅ d x d p + 2 p a − 2 p 2 a x ⋅ d x d p d p d x ( x 2 − a x 2 p 2 ) = p 2 − a 2 p \frac {d p}{d x} \left(\frac {x}{2} - \frac {a x}{2 p ^ {2}}\right) = \frac {p}{2} - \frac {a}{2 p} d x d p ( 2 x − 2 p 2 a x ) = 2 p − 2 p a x d p d x ( 1 2 − a 2 p 2 ) = p ( 1 2 − a 2 p 2 ) x \frac {d p}{d x} \left(\frac {1}{2} - \frac {a}{2 p ^ {2}}\right) = p \left(\frac {1}{2} - \frac {a}{2 p ^ {2}}\right) x d x d p ( 2 1 − 2 p 2 a ) = p ( 2 1 − 2 p 2 a ) x d p d x = p , 1 2 − a 2 p 2 ≠ 0 x \frac {d p}{d x} = p, \frac {1}{2} - \frac {a}{2 p ^ {2}} \neq 0 x d x d p = p , 2 1 − 2 p 2 a = 0 d p p = d x x \frac {d p}{p} = \frac {d x}{x} p d p = x d x
Integrate
ln ∣ p ∣ = ln ∣ x ∣ + ln c \ln | p | = \ln | x | + \ln c ln ∣ p ∣ = ln ∣ x ∣ + ln c p = c x p = c x p = c x
Substitute p = c x p = cx p = c x ( ∗ ) (\ast) ( ∗ )
x ( c x ) 2 − 2 y ( c x ) + a x = 0 x (c x) ^ {2} - 2 y (c x) + a x = 0 x ( c x ) 2 − 2 y ( c x ) + a x = 0 c 2 x 3 + x ( − 2 y c + a ) = 0 ⇒ c 2 x 2 − 2 y c + a = 0 ⇒ 2 y = c x 2 + a c c ^ {2} x ^ {3} + x (- 2 y c + a) = 0 \Rightarrow c ^ {2} x ^ {2} - 2 y c + a = 0 \Rightarrow 2 y = c x ^ {2} + \frac {a}{c} c 2 x 3 + x ( − 2 yc + a ) = 0 ⇒ c 2 x 2 − 2 yc + a = 0 ⇒ 2 y = c x 2 + c a
The general solution of the differential equation
2 y = c x 2 + a c 2 y = c x ^ {2} + \frac {a}{c} 2 y = c x 2 + c a
If 1 2 − a 2 p 2 = 0 \frac{1}{2} -\frac{a}{2p^2} = 0 2 1 − 2 p 2 a = 0 p = − a p = -\sqrt{a} p = − a p = a p = \sqrt{a} p = a
Substitute p = − a p = -\sqrt{a} p = − a ( ∗ ) (\ast) ( ∗ )
a x + 2 a y + a x = 0 y = − a x \begin{array}{l}
ax + 2\sqrt{a}y + ax = 0 \\
y = -\sqrt{a}x
\end{array} a x + 2 a y + a x = 0 y = − a x
Substitute p = a p = \sqrt{a} p = a ( ∗ ) (\ast) ( ∗ )
a x − 2 a y + a x = 0 y = a x \begin{array}{l}
ax - 2\sqrt{a}y + ax = 0 \\
y = \sqrt{a}x
\end{array} a x − 2 a y + a x = 0 y = a x
The solutions of the given differential equation are
2 y = c x 2 + a c , or y = − a x , or y = a x 2y = c x^2 + \frac{a}{c}, \text{ or } y = -\sqrt{a}x, \text{ or } y = \sqrt{a}x 2 y = c x 2 + c a , or y = − a x , or y = a x
Answer: 2 y = c x 2 + a c , y = − a x , y = a x . 2y = c x^2 + \frac{a}{c}, y = -\sqrt{a}x, y = \sqrt{a}x. 2 y = c x 2 + c a , y = − a x , y = a x .
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#340153 on Dec 2023