Answer on Question #58422 – Math – Differential Equations
Question
6. Solve
( x 3 + y 3 ) d x − 3 x y 2 d y = 0 (x^3 + y^3)dx - 3xy^2dy = 0 ( x 3 + y 3 ) d x − 3 x y 2 d y = 0
Solution
( x 3 + y 3 ) d x = 3 x y 2 d y , x 3 + y 3 3 x y 2 = d y d x , y ′ = 1 3 ( x 2 y 2 + y x ) (x^3 + y^3)dx = 3xy^2dy, \quad \frac{x^3 + y^3}{3xy^2} = \frac{dy}{dx}, \quad y' = \frac{1}{3}\left(\frac{x^2}{y^2} + \frac{y}{x}\right) ( x 3 + y 3 ) d x = 3 x y 2 d y , 3 x y 2 x 3 + y 3 = d x d y , y ′ = 3 1 ( y 2 x 2 + x y )
Substitution y = t x y = tx y = t x t = t ( x ) t = t(x) t = t ( x ) y ′ = t + t ′ x y' = t + t'x y ′ = t + t ′ x t = y x t = \frac{y}{x} t = x y
We obtain t + t ′ x = 1 3 ( 1 t 2 + t ) t + t'x = \frac{1}{3}\left(\frac{1}{t^2} + t\right) t + t ′ x = 3 1 ( t 2 1 + t ) t ′ x = 1 3 ( 1 t 2 − 2 t ) = 1 − 2 t 3 3 t 2 t'x = \frac{1}{3}\left(\frac{1}{t^2} - 2t\right) = \frac{1 - 2t^3}{3t^2} t ′ x = 3 1 ( t 2 1 − 2 t ) = 3 t 2 1 − 2 t 3 d t d x x = 1 − 2 t 3 3 t 2 \frac{dt}{dx}x = \frac{1 - 2t^3}{3t^2} d x d t x = 3 t 2 1 − 2 t 3 3 t 2 d t ( 1 − 2 t 3 ) = d x x \frac{3t^2dt}{(1 - 2t^3)} = \frac{dx}{x} ( 1 − 2 t 3 ) 3 t 2 d t = x d x
∫ 3 t 2 d t ( 1 − 2 t 3 ) = ∫ d x x , − 1 2 ln ∣ 1 − 2 t 3 ∣ = ln ∣ C x ∣ , 1 − 2 t 3 = C x 2 , 1 − 2 y 3 x 3 = C x 2 , \int \frac{3t^2dt}{(1 - 2t^3)} = \int \frac{dx}{x}, \quad -\frac{1}{2}\ln |1 - 2t^3| = \ln |Cx|, \quad 1 - 2t^3 = \frac{C}{x^2}, \quad 1 - 2\frac{y^3}{x^3} = \frac{C}{x^2}, ∫ ( 1 − 2 t 3 ) 3 t 2 d t = ∫ x d x , − 2 1 ln ∣1 − 2 t 3 ∣ = ln ∣ C x ∣ , 1 − 2 t 3 = x 2 C , 1 − 2 x 3 y 3 = x 2 C , x 3 − 2 y 3 = C x x^3 - 2y^3 = Cx x 3 − 2 y 3 = C x
Answer: x 3 − 2 y 3 = C x x^3 - 2y^3 = Cx x 3 − 2 y 3 = C x
Question
7. Solve
( 1 + 2 e x y ) d x + 2 e x y ( 1 − x y ) d y = 0 (1 + 2e^{xy})dx + 2e^{xy}(1 - xy)dy = 0 ( 1 + 2 e x y ) d x + 2 e x y ( 1 − x y ) d y = 0 5 x + 2 y 5x + 2y 5 x + 2 y
x + 2 y x + 2y x + 2 y
x + 2 y x + 2y x + 2 y
5 x + 3 y 5x + 3y 5 x + 3 y
(May be a bug)
Solution
According to the suggested answer options, this equation must be the total differential equation, or must become it through the introduction of some integrating factors. But we have:
d ( 5 x + 2 y e x y ) = ( 5 + 2 y 2 e x y ) d x + 2 e x y ( 1 + x y ) d y d ( x + 2 y e 2 x y ) = ( 1 + 4 y 2 e 2 x y ) d x + 2 e 2 x y ( 1 + 2 x y ) d y d ( x + 2 y e x y ) = ( 1 + 2 y 2 e x y ) d x + 2 e x y ( 1 + x y ) d y d ( 5 x + 3 y e x y ) = ( 5 + 3 y 2 e x y ) d x + 3 e x y ( 1 + x y ) d y \begin{array}{l}
d(5x + 2ye^{xy}) = (5 + 2y^2e^{xy})dx + 2e^{xy}(1 + xy)dy \\
d(x + 2ye^{2xy}) = (1 + 4y^2e^{2xy})dx + 2e^{2xy}(1 + 2xy)dy \\
d(x + 2ye^{xy}) = (1 + 2y^2e^{xy})dx + 2e^{xy}(1 + xy)dy \\
d(5x + 3ye^{xy}) = (5 + 3y^2e^{xy})dx + 3e^{xy}(1 + xy)dy \\
\end{array} d ( 5 x + 2 y e x y ) = ( 5 + 2 y 2 e x y ) d x + 2 e x y ( 1 + x y ) d y d ( x + 2 y e 2 x y ) = ( 1 + 4 y 2 e 2 x y ) d x + 2 e 2 x y ( 1 + 2 x y ) d y d ( x + 2 y e x y ) = ( 1 + 2 y 2 e x y ) d x + 2 e x y ( 1 + x y ) d y d ( 5 x + 3 y e x y ) = ( 5 + 3 y 2 e x y ) d x + 3 e x y ( 1 + x y ) d y
Answer: none of them.
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
d y d x = − y ( x y + 1 ) x ( 1 + x y + x 2 y 2 ) \frac{dy}{dx} = -\frac{y(xy + 1)}{x(1 + xy + x^2y^2)} d x d y = − x ( 1 + x y + x 2 y 2 ) y ( x y + 1 )
Substitution x y = t xy = t x y = t t = t ( x ) t = t(x) t = t ( x ) y = t x y = \frac{t}{x} y = x t y ′ = t ′ x − t x 2 = t ′ x − t x 2 y' = \frac{t'x - t}{x^2} = \frac{t'}{x} - \frac{t}{x^2} y ′ = x 2 t ′ x − t = x t ′ − x 2 t
We obtain
t ′ x − t x 2 = − t ( t + 1 ) x 2 ( 1 + t + t 2 ) ; t ′ x − t = − t ( t + 1 ) ( 1 + t + t 2 ) ; t ′ x = t ( 1 − t + 1 1 + t + t 2 ) ; d t d x x = t 3 1 + t + t 2 ; ( 1 + t + t 2 ) d t t 3 = d x x ; \begin{array}{l}
\frac{t'}{x} - \frac{t}{x^2} = -\frac{t(t + 1)}{x^2(1 + t + t^2)}; \quad t'x - t = -\frac{t(t + 1)}{(1 + t + t^2)}; \quad t'x = t\left(1 - \frac{t + 1}{1 + t + t^2}\right); \\
\frac{dt}{dx}x = \frac{t^3}{1 + t + t^2}; \quad \frac{(1 + t + t^2)dt}{t^3} = \frac{dx}{x}; \\
\end{array} x t ′ − x 2 t = − x 2 ( 1 + t + t 2 ) t ( t + 1 ) ; t ′ x − t = − ( 1 + t + t 2 ) t ( t + 1 ) ; t ′ x = t ( 1 − 1 + t + t 2 t + 1 ) ; d x d t x = 1 + t + t 2 t 3 ; t 3 ( 1 + t + t 2 ) d t = x d x ; ∫ ( 1 + t + t 2 ) d t t 3 = ∫ d x x , − 1 2 t 2 − 1 t + ln ∣ t ∣ = ln ∣ x ∣ + C , − 1 2 t 2 − 1 t + ln ∣ t x ∣ = C \int \frac{(1 + t + t^2)dt}{t^3} = \int \frac{dx}{x}, \quad -\frac{1}{2t^2} - \frac{1}{t} + \ln |t| = \ln |x| + C, \quad -\frac{1}{2t^2} - \frac{1}{t} + \ln \left| \frac{t}{x} \right| = C ∫ t 3 ( 1 + t + t 2 ) d t = ∫ x d x , − 2 t 2 1 − t 1 + ln ∣ t ∣ = ln ∣ x ∣ + C , − 2 t 2 1 − t 1 + ln ∣ ∣ x t ∣ ∣ = C 2 t 2 ln ∣ t x ∣ − 2 t − 1 = t 2 C , 2t^2 \ln \left| \frac{t}{x} \right| - 2t - 1 = t^2 C, 2 t 2 ln ∣ ∣ x t ∣ ∣ − 2 t − 1 = t 2 C , 2 x 2 y 2 ln ∣ y ∣ − 2 x y − 1 = x 2 y 2 C 2x^2 y^2 \ln |y| - 2xy - 1 = x^2 y^2 C 2 x 2 y 2 ln ∣ y ∣ − 2 x y − 1 = x 2 y 2 C
Answer: 2 x 2 y 2 ln ∣ y ∣ − 2 x y − 1 = C x 2 y 2 2x^2 y^2 \ln |y| - 2xy - 1 = Cx^2 y^2 2 x 2 y 2 ln ∣ y ∣ − 2 x y − 1 = C x 2 y 2
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Question
9. Solve
x d y − y d x − x 2 − y 2 − d x = 0 xdy - ydx - x^2 - y^2 - \sqrt{dx} = 0 x d y − y d x − x 2 − y 2 − d x = 0 C x = 2 e arcsin y x Cx = 2e \arcsin yx C x = 2 e arcsin y x C x = e arcsin y x Cx = e \arcsin yx C x = e arcsin y x C x = e arcsin 2 y 3 x Cx = e \arcsin^2 y^3 x C x = e arcsin 2 y 3 x C x = e arccos y x Cx = e \arccos yx C x = e arccos y x
Answer: I think that the statement of this question is incorrect.
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Question
10. The population of student P at NOUN increases at a rate proportional to the population and to the addition of 150,250 and the population divided by 3, the differential equation of this statement is
Solution.
d P d T = k P ( 150 , 250 + P ) : 3 , \frac{dP}{dT} = kP(150,250 + P): 3, d T d P = k P ( 150 , 250 + P ) : 3 ,
Answer: d P d T = k P ( 150 , 250 + P ) : 3 \frac{dP}{dT} = kP(150,250 + P): 3 d T d P = k P ( 150 , 250 + P ) : 3
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#339914 on Dec 2023