ANSWER on Question #79371 – Math – Differential Equations
QUESTION
Solve the differential equation
d y d x = 2 y 2 + 3 x y / x 2 \frac{dy}{dx} = 2y^2 + 3xy/x^2 d x d y = 2 y 2 + 3 x y / x 2
Possible answers:
a)
y = c x 3 1 − c x 2 y = \frac{cx^3}{1 - cx^2} y = 1 − c x 2 c x 3
b)
y = c x 31 + c x 2 y = cx31 + cx2 y = c x 31 + c x 2
c)
y = − c x 31 − c x 2 y = -cx31 - cx2 y = − c x 31 − c x 2
d)
y = − c x 31 + c x 2 y = -cx31 + cx2 y = − c x 31 + c x 2 SOLUTION
Hint: Since only the version of the answer (a) is written with the help of Latex functions, only its formula editor understands uniquely, only I will consider it correct from the whole question.
Hint: Equation Editor predetermined equation may be understood in two ways
1)
d y d x = 2 y 2 + 3 x y x 2 \frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} d x d y = x 2 2 y 2 + 3 x y
2)
d y d x = 2 y 2 + 3 x y x 2 \frac{dy}{dx} = 2y^2 + \frac{3xy}{x^2} d x d y = 2 y 2 + x 2 3 x y
We will solve each of these equations separately.
*Hint:* Most likely, in the variants of the answer some function signs are omitted. Probably, the answer options should have looked like this
a)
y = c x 3 1 − c x 2 y = \frac{cx^3}{1 - cx^2} y = 1 − c x 2 c x 3
b)
y = c x 3 1 + c x 2 y = \frac{cx^3}{1 + cx^2} y = 1 + c x 2 c x 3
c)
y = − c x 3 1 − c x 2 y = -\frac{cx^3}{1 - cx^2} y = − 1 − c x 2 c x 3
d)
y = − c x 3 1 + c x 2 y = -\frac{cx^3}{1 + cx^2} y = − 1 + c x 2 c x 3
Now we come to the solution of this equation.
1 CASE:
d y d x = 2 y 2 + 3 x y x 2 → d y d x = 2 y 2 x 2 + 3 x y x 2 → d y d x = 2 ⋅ ( y x ) 2 + 3 ⋅ y x \frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} \rightarrow \frac{dy}{dx} = \frac{2y^2}{x^2} + \frac{3xy}{x^2} \rightarrow \boxed{\frac{dy}{dx} = 2 \cdot \left(\frac{y}{x}\right)^2 + 3 \cdot \frac{y}{x}} d x d y = x 2 2 y 2 + 3 x y → d x d y = x 2 2 y 2 + x 2 3 x y → d x d y = 2 ⋅ ( x y ) 2 + 3 ⋅ x y
We introduce the substitution
u = y x → y = u x → d y d x = d u d x ⋅ x + u ⋅ 1 → d y d x = d u d x ⋅ x + u u = \frac{y}{x} \rightarrow y = ux \rightarrow \frac{dy}{dx} = \frac{du}{dx} \cdot x + u \cdot 1 \rightarrow \boxed{\frac{dy}{dx} = \frac{du}{dx} \cdot x + u} u = x y → y = ux → d x d y = d x d u ⋅ x + u ⋅ 1 → d x d y = d x d u ⋅ x + u
Then
{ d y d x = 2 ⋅ ( y x ) 2 + 3 ⋅ y x u = y x d y d x = d u d x ⋅ x + u → d u d x ⋅ x + u = 2 u 2 + 3 u → d u d x ⋅ x = 2 u 2 + 3 u − u → d u d x ⋅ x = 2 u 2 + 2 u → \begin{array}{l}
\left\{
\begin{array}{c}
\frac{dy}{dx} = 2 \cdot \left(\frac{y}{x}\right)^2 + 3 \cdot \frac{y}{x} \\
\quad u = \frac{y}{x} \\
\quad
\frac{dy}{dx} = \frac{du}{dx} \cdot x + u
\end{array}
\right.
\rightarrow
\frac{du}{dx} \cdot x + u = 2u^2 + 3u \rightarrow \frac{du}{dx} \cdot x = 2u^2 + 3u - u \rightarrow \frac{du}{dx} \cdot x \\
= 2u^2 + 2u \rightarrow \\
\end{array} ⎩ ⎨ ⎧ d x d y = 2 ⋅ ( x y ) 2 + 3 ⋅ x y u = x y d x d y = d x d u ⋅ x + u → d x d u ⋅ x + u = 2 u 2 + 3 u → d x d u ⋅ x = 2 u 2 + 3 u − u → d x d u ⋅ x = 2 u 2 + 2 u → d u d x ⋅ x = 2 u 2 + 2 u ∣ × ( 2 ⋅ d x x ⋅ ( 2 u 2 + 2 u ) ) → 2 ⋅ d u 2 u 2 + 2 u = 2 ⋅ d x x → 2 ⋅ ( ( u + 1 ) − u ) d u 2 u ( u + 1 ) = 2 ⋅ d x x → \begin{array}{l}
\frac{du}{dx} \cdot x = 2u^2 + 2u \Bigg| \times \left(\frac{2 \cdot dx}{x \cdot (2u^2 + 2u)}\right) \rightarrow \frac{2 \cdot du}{2u^2 + 2u} = \frac{2 \cdot dx}{x} \rightarrow \frac{2 \cdot ((u + 1) - u)du}{2u(u + 1)} \\
= \frac{2 \cdot dx}{x} \rightarrow \\
\end{array} d x d u ⋅ x = 2 u 2 + 2 u ∣ ∣ × ( x ⋅ ( 2 u 2 + 2 u ) 2 ⋅ d x ) → 2 u 2 + 2 u 2 ⋅ d u = x 2 ⋅ d x → 2 u ( u + 1 ) 2 ⋅ (( u + 1 ) − u ) d u = x 2 ⋅ d x → ( 1 u − 1 u + 1 ) d u = ( 2 x ) d x → ∫ ( 1 u − 1 u + 1 ) d u = ∫ ( 2 x ) d x → ln ∣ u ∣ − ln ∣ u + 1 ∣ = 2 ⋅ ln ∣ x ∣ + ln ∣ c ∣ → \begin{array}{l}
\left(\frac{1}{u} - \frac{1}{u + 1}\right) du = \left(\frac{2}{x}\right) dx \rightarrow \int \left(\frac{1}{u} - \frac{1}{u + 1}\right) du = \int \left(\frac{2}{x}\right) dx \rightarrow \ln |u| - \ln |u + 1| \\
= 2 \cdot \ln |x| + \ln |c| \rightarrow \\
\end{array} ( u 1 − u + 1 1 ) d u = ( x 2 ) d x → ∫ ( u 1 − u + 1 1 ) d u = ∫ ( x 2 ) d x → ln ∣ u ∣ − ln ∣ u + 1∣ = 2 ⋅ ln ∣ x ∣ + ln ∣ c ∣ → ln ∣ u u + 1 ∣ = ln ∣ c x 2 ∣ → u u + 1 = c x 2 ∣ × ( u + 1 ) → u = c x 2 ⋅ ( u + 1 ) → u = c x 2 ⋅ u + c x 2 → \ln \left| \frac{u}{u + 1} \right| = \ln |cx^2| \rightarrow \frac{u}{u + 1} = cx^2 \Bigg| \times (u + 1) \rightarrow u = cx^2 \cdot (u + 1) \rightarrow u = cx^2 \cdot u + cx^2 \rightarrow ln ∣ ∣ u + 1 u ∣ ∣ = ln ∣ c x 2 ∣ → u + 1 u = c x 2 ∣ ∣ × ( u + 1 ) → u = c x 2 ⋅ ( u + 1 ) → u = c x 2 ⋅ u + c x 2 → u − c x 2 ⋅ u = c x 2 → u ( 1 − c x 2 ) = c x 2 → u = c x 2 1 − c x 2 u - cx^2 \cdot u = cx^2 \rightarrow u(1 - cx^2) = cx^2 \rightarrow \boxed{u = \frac{cx^2}{1 - cx^2}} u − c x 2 ⋅ u = c x 2 → u ( 1 − c x 2 ) = c x 2 → u = 1 − c x 2 c x 2
We recall that we introduced a substitution
u = y x u = \frac{y}{x} u = x y
Then,
{ u = c x 2 1 − c x 2 → y x = c x 2 1 − c x 2 u = y x × ( x ) → y = c x 3 1 − c x 2 \left\{
\begin{array}{c}
u = \frac{cx^2}{1 - cx^2} \rightarrow \frac{y}{x} = \frac{cx^2}{1 - cx^2} \\
\quad u = \frac{y}{x}
\end{array}
\right.
\times (x) \rightarrow y = \frac{cx^3}{1 - cx^2} { u = 1 − c x 2 c x 2 → x y = 1 − c x 2 c x 2 u = x y × ( x ) → y = 1 − c x 2 c x 3
Conclusion,
d y d x = 2 y 2 + 3 x y x 2 → y = c x 3 1 − c x 2 \boxed{\frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} \rightarrow y = \frac{cx^3}{1 - cx^2}} d x d y = x 2 2 y 2 + 3 x y → y = 1 − c x 2 c x 3
2 CASE:
d y d x = 2 y 2 + 3 x y x 2 → d y d x = 2 y 2 + 3 ⋅ y x \frac {d y}{d x} = 2 y ^ {2} + \frac {3 x y}{x ^ {2}} \rightarrow \frac {d y}{d x} = 2 y ^ {2} + 3 \cdot \frac {y}{x} d x d y = 2 y 2 + x 2 3 x y → d x d y = 2 y 2 + 3 ⋅ x y
As we can see, this is the Bernoulli equation:
d y d x + P ( x ) y = Q ( x ) y n \frac {d y}{d x} + P (x) y = Q (x) y ^ {n} d x d y + P ( x ) y = Q ( x ) y n
(More information: https://en.wikipedia.org/wiki/Bernoulli_differential_equation)
In our case,
d y d x = 2 y 2 + 3 ⋅ y x ∣ ÷ ( y 2 ) → 1 y 2 ⋅ d y d x = 2 + 3 x ⋅ 1 y \frac {d y}{d x} = 2 y ^ {2} + 3 \cdot \frac {y}{x} \mid \div (y ^ {2}) \rightarrow \frac {1}{y ^ {2}} \cdot \frac {d y}{d x} = 2 + \frac {3}{x} \cdot \frac {1}{y} d x d y = 2 y 2 + 3 ⋅ x y ∣ ÷ ( y 2 ) → y 2 1 ⋅ d x d y = 2 + x 3 ⋅ y 1
We introduce the substitution
u = 1 y → d u d x = − 1 y 2 ⋅ d y d x → { y = 1 u 1 y 2 ⋅ d y d x = − d u d x ] u = \frac {1}{y} \rightarrow \frac {d u}{d x} = - \frac {1}{y ^ {2}} \cdot \frac {d y}{d x} \rightarrow \boxed {\left\{\begin{array}{c}y = \frac {1}{u}\\\frac {1}{y ^ {2}} \cdot \frac {d y}{d x} = - \frac {d u}{d x}\end{array}\right]} u = y 1 → d x d u = − y 2 1 ⋅ d x d y → { y = u 1 y 2 1 ⋅ d x d y = − d x d u ]
Then,
{ 1 y 2 ⋅ d y d x = 2 + 3 x ⋅ 1 y y = 1 u 1 y 2 ⋅ d y d x = − d u d x → − d u d x = 2 + 3 x ⋅ u \left\{\begin{array}{c}\frac {1}{y ^ {2}} \cdot \frac {d y}{d x} = 2 + \frac {3}{x} \cdot \frac {1}{y}\\\quad y = \frac {1}{u}\\\frac {1}{y ^ {2}} \cdot \frac {d y}{d x} = - \frac {d u}{d x}\end{array}\right. \rightarrow \boxed {- \frac {d u}{d x} = 2 + \frac {3}{x} \cdot u} ⎩ ⎨ ⎧ y 2 1 ⋅ d x d y = 2 + x 3 ⋅ y 1 y = u 1 y 2 1 ⋅ d x d y = − d x d u → − d x d u = 2 + x 3 ⋅ u
We have obtained a nonhomogeneous equation of the first order. We solve it by the method of variation of the constant.
1 STEP: Solve a homogeneous equation.
− d u d x = 3 x ⋅ u ∣ × ( − d x u ) → d u u = − 3 ⋅ d x x → ∫ ( 1 u ) d u = ∫ ( − 3 x ) d x → ln ∣ u ∣ = − 3 ⋅ ln ∣ x ∣ + ln ∣ c ∣ → \begin{array}{l} - \frac {d u}{d x} = \frac {3}{x} \cdot u \Bigg | \times \left(\frac {- d x}{u}\right) \rightarrow \frac {d u}{u} = \frac {- 3 \cdot d x}{x} \rightarrow \int \left(\frac {1}{u}\right) d u = \int \left(- \frac {3}{x}\right) d x \rightarrow \ln | u | \\ = - 3 \cdot \ln | x | + \ln | c | \rightarrow \\ \end{array} − d x d u = x 3 ⋅ u ∣ ∣ × ( u − d x ) → u d u = x − 3 ⋅ d x → ∫ ( u 1 ) d u = ∫ ( − x 3 ) d x → ln ∣ u ∣ = − 3 ⋅ ln ∣ x ∣ + ln ∣ c ∣ → ln ∣ u ∣ = ln ∣ c x 3 ∣ → u = c x 3 \ln | u | = \ln \left| \frac {c}{x ^ {3}} \right| \rightarrow \boxed {u = \frac {c}{x ^ {3}}} ln ∣ u ∣ = ln ∣ ∣ x 3 c ∣ ∣ → u = x 3 c
2 STEP: Variations of the constant.
We will assume that c = c ( x ) c = c(x) c = c ( x )
u = c ( x ) x 3 → d u d x = 1 x 3 ⋅ d c d x − 3 c ( x ) x 4 u = \frac {c (x)}{x ^ {3}} \rightarrow \boxed {\frac {d u}{d x} = \frac {1}{x ^ {3}} \cdot \frac {d c}{d x} - \frac {3 c (x)}{x ^ {4}}} u = x 3 c ( x ) → d x d u = x 3 1 ⋅ d x d c − x 4 3 c ( x )
We substitute the obtained formula for the derivative in the initial equation
\left\{ \begin{array}{l} - \frac {d u}{d x} = 2 + \frac {3}{x} \cdot u \\ u = \frac {c (x)}{x ^ {3}} \\ \frac {d u}{d x} = \frac {1}{x ^ {3}} \cdot \frac {d c}{d x} - \frac {3 c}{x ^ {4}} \end{array} \right. \to - \left(\frac {1}{x ^ {3}} \cdot \frac {d c}{d x} - \frac {3 c (x)}{x ^ {4}}\right) = 2 + \frac {3}{x} \cdot \frac {c (x)}{x ^ {3}} \to \\ - \frac {1}{x ^ {3}} \cdot \frac {d c}{d x} + \frac {3 c (x)}{x ^ {4}} = 2 + \frac {3 c (x)}{x ^ {4}} \to - \frac {1}{x ^ {3}} \cdot \frac {d c}{d x} = 2 \Bigg | \times (- x ^ {3} d x) \to d c = (- 2 x ^ {3}) d x \to \\ \end{array} \right.
Then,
{ u = c ( x ) x 3 c ( x ) = A − x 4 2 → u = A − x 4 2 x 3 \left\{ \begin{array}{c} u = \frac {c (x)}{x ^ {3}} \\ c (x) = \frac {A - x ^ {4}}{2} \end{array} \to \boxed {u = \frac {A - x ^ {4}}{2 x ^ {3}}} \right. { u = x 3 c ( x ) c ( x ) = 2 A − x 4 → u = 2 x 3 A − x 4
We recall that we introduced a substitution
u = 1 y u = \frac {1}{y} u = y 1
Then
{ u = A − x 4 2 x 3 u = 1 y → 1 y = A − x 4 2 x 3 → y = 2 x 3 A − x 4 \left\{ \begin{array}{c} u = \frac {A - x ^ {4}}{2 x ^ {3}} \\ u = \frac {1}{y} \end{array} \right. \to \frac {1}{y} = \frac {A - x ^ {4}}{2 x ^ {3}} \to y = \frac {2 x ^ {3}}{A - x ^ {4}} { u = 2 x 3 A − x 4 u = y 1 → y 1 = 2 x 3 A − x 4 → y = A − x 4 2 x 3
Conclusion,
d y d x = 2 y 2 + 3 x y x 2 → y = 2 x 3 A − x 4 \boxed {\frac {d y}{d x} = 2 y ^ {2} + \frac {3 x y}{x ^ {2}} \rightarrow y = \frac {2 x ^ {3}}{A - x ^ {4}}} d x d y = 2 y 2 + x 2 3 x y → y = A − x 4 2 x 3 ANSWER:
If the initial equation was
d y d x = 2 y 2 + 3 x y x 2 , \frac {d y}{d x} = \frac {2 y ^ {2} + 3 x y}{x ^ {2}}, d x d y = x 2 2 y 2 + 3 x y ,
then the solution is
a ) y = c x 3 1 − c x 2 a) y = \frac {c x ^ {3}}{1 - c x ^ {2}} a ) y = 1 − c x 2 c x 3
If the initial equation was
d y d x = 2 y 2 + 3 x y x 2 \frac {d y}{d x} = 2 y ^ {2} + \frac {3 x y}{x ^ {2}} d x d y = 2 y 2 + x 2 3 x y
then
In the provided options there is no correct answer
d y d x = 2 y 2 + 3 x y x 2 → y = 2 x 3 A − x 4 will be the correct answer . \frac {d y}{d x} = 2 y ^ {2} + \frac {3 x y}{x ^ {2}} \rightarrow y = \frac {2 x ^ {3}}{A - x ^ {4}} \text{ will be the correct answer}. d x d y = 2 y 2 + x 2 3 x y → y = A − x 4 2 x 3 will be the correct answer .
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#340153 on Dec 2023