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

Recognize Separable Equation: Recognize that the given differential equation is separable, meaning we can separate the variables y and x on different sides of the equation. (dy)/(dx) = xe^(-y) + 10e^(-y) We can write this as e^(y)dy = (x + 10)dx.Integrate with Respect: Integrate both sides of the equation with respect to their respective variables. ∫e^(y)dy = ∫(x + 10)dxPerform Integration: Perform the integration. The integral of e^(y) with respect to y is e^(y), and the integral of (x + 10) with respect to x is (x^2)/2 + 10x. So we have e^(y) = (x^2)/2 + 10x + C, where C is the constant of integration.Solve for y: Solve for y by taking the natural logarithm of both sides. y = ln((x^2)/2 + 10x + C)Check Answer Choices: Check the answer choices to see which one matches our solution. The correct answer is (C) y = ln((x^2)/2 + 10x + C).

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Solve the equation.

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

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

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Find derivatives of using multiple formulae

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Q. Solve the equation.

\newline

\frac{d y}{d x}=x e^{-y}+10 e^{-y}

\newline

Choose

1

answer:

\newline

(A)

y=\frac{x^{2}}{2}+10 x+C

\newline

(B)

y=-\ln \left(-\frac{x^{2}}{2}-10 x+C\right)

\newline

(C)

y=\ln \left(\frac{x^{2}}{2}+10 x+C\right)

\newline

(D)

y=-\ln \left(-\frac{x^{2}}{2}-10 x\right)+C

Recognize Separable Equation: Recognize that the given differential equation is separable, meaning we can separate the variables $y$ and $x$ on different sides of the equation. $\newline$ $\frac{dy}{dx} = x e^{-y} + 10 e^{-y}$ $\newline$ We can write this as $e^{y}dy = (x + 10)dx$ .
Integrate with Respect: Integrate both sides of the equation with respect to their respective variables. $\newline$ $\int e^{y} \, dy = \int (x + 10) \, dx$
Perform Integration: Perform the integration. $\newline$ The integral of $e^{y}$ with respect to $y$ is $e^{y}$ , and the integral of $(x + 10)$ with respect to $x$ is $(x^2)/2 + 10x$ . $\newline$ So we have $e^{y} = (x^2)/2 + 10x + C$ , where $C$ is the constant of integration.
Solve for y: Solve for y by taking the natural logarithm of both sides. $\newline$ $y = \ln(\frac{x^2}{2} + 10x + C)$
Check Answer Choices: Check the answer choices to see which one matches our solution. $\newline$ The correct answer is (C) $y = \ln\left(\frac{x^2}{2} + 10x + C\right)$ .