Solve the equation. (dy)/(dx)=-10x^(4)y Choose 1 answer: (A) y=-2(x+C)^(5) (B) y=Ce^(-2x^(5)) (C) y=-2x^(5)+C (D) y=e^(-2x^(5))+C

Separate variables: We are given a first-order separable differential equation (dy)/(dx)=-10x^(4)y. To solve it, we need to separate the variables y and x on different sides of the equation.Rewrite equation: Rewrite the equation to separate the variables: (1/y)dy = -10x^(4)dx.Integrate both sides: Integrate both sides of the equation. The left side with respect to y and the right side with respect to x. ∫(1/y)dy = ∫-10x^(4)dx.Exponentiate both sides: The integral of (1/y)dy is ln|y|, and the integral of -10x^(4)dx is -2x^(5). Don't forget to add the constant of integration C on one side. ln|y| = -2x^(5) + C.Solve for y: To solve for y, we exponentiate both sides of the equation to get rid of the natural logarithm. e^(ln|y|) = e^(-2x^(5) + C).Final solution: Since e^(ln|y|) is just |y| and e^(C) is a constant which we can call C', we have: |y| = C'e^(-2x^(5)).Since y can be positive or negative, we can drop the absolute value to get: y = ±C'e^(-2x^(5)). However, since the constant C' can absorb the ± sign, we can write: y = Ce^(-2x^(5)).

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Solve the equation.

(dy)/(dx)=-10x^(4)y
Choose 1 answer:
(A)
y=-2(x+C)^(5)
(B)
y=Ce^(-2x^(5))
(C)
y=-2x^(5)+C
(D)
y=e^(-2x^(5))+C

Solve the equation. $\newline$ $\frac{d y}{d x}=-10 x^{4} y$ $\newline$ Choose $1$ answer: $\newline$ (A) $y=-2(x+C)^{5}$ $\newline$ (B) $y=C e^{-2 x^{5}}$ $\newline$ (C) $y=-2 x^{5}+C$ $\newline$ (D) $y=e^{-2 x^{5}}+C$

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Q. Solve the equation.

\newline

\frac{d y}{d x}=-10 x^{4} y

\newline

Choose

1

answer:

\newline

(A)

y=-2(x+C)^{5}

\newline

(B)

y=C e^{-2 x^{5}}

\newline

(C)

y=-2 x^{5}+C

\newline

(D)

y=e^{-2 x^{5}}+C

Separate variables: We are given a first-order separable differential equation $(\frac{dy}{dx})=-10x^{4}y$ . To solve it, we need to separate the variables $y$ and $x$ on different sides of the equation.
Rewrite equation: Rewrite the equation to separate the variables: $(\frac{1}{y})dy = -10x^{4}dx$ .
Integrate both sides: Integrate both sides of the equation. The left side with respect to $y$ and the right side with respect to $x$ . $\int(\frac{1}{y})dy = \int-10x^{4}dx.$
Exponentiate both sides: The integral of $\frac{1}{y}$ dy is $\ln|y|$ , and the integral of $-10x^{4}$ dx is $-2x^{5}$ . Don't forget to add the constant of integration $C$ on one side. $\newline$ $\ln|y| = -2x^{5} + C$ .
Solve for $y$ : To solve for $y$ , we exponentiate both sides of the equation to get rid of the natural logarithm. $e^{(\ln|y|)} = e^{(-2x^{5} + C)}.$
Final solution: Since $e^{\ln|y|}$ is just $|y|$ and $e^{C}$ is a constant which we can call $C'$ , we have: $\newline$ $|y| = C'e^{-2x^{5}}$ .
Final solution: Since $e^{\ln|y|}$ is just $|y|$ and $e^{C}$ is a constant which we can call $C'$ , we have: $\newline$ $|y| = C'e^{-2x^{5}}$ .Since $y$ can be positive or negative, we can drop the absolute value to get: $\newline$ $y = \pm C'e^{-2x^{5}}$ . $\newline$ However, since the constant $C'$ can absorb the $\pm$ sign, we can write: $\newline$ $y = Ce^{-2x^{5}}$ .