Solve the equation. (dy)/(dx)=-(1)/(4)e^(x)y^(-2) Choose 1 answer: (A) y=+-root(3)(-(3e^(x))/(4)+C) (B) y=root(3)(-(3e^(x))/(4)+C) (C) y=+-root(3)(-(3e^(x))/(4))+C (D) y=root(3)(-(3e^(x))/(4))+C

Separate variables: We are given the differential equation (dy)/(dx) = -(1/4)e^(x)y^(-2). This is a separable differential equation, which means we can separate the variables y and x on different sides of the equation.Multiply and integrate: To separate the variables, we multiply both sides by y^2 and dx to get y^2 dy = -(1/4)e^(x) dx.Integrate both sides: Now we integrate both sides of the equation. The left side with respect to y, and the right side with respect to x. ∫y^2 dy = ∫-(1/4)e^(x) dxSimplify and rename constant: The integral of y^2 with respect to y is (1/3)y^3. The integral of -(1/4)e^(x) with respect to x is -(1/4)e^(x). So we have: (1/3)y^3 = -(1/4)e^(x) + C, where C is the constant of integration.Take cube root: We multiply through by 3 to get rid of the fraction on the left side: y^3 = -(3/4)e^(x) + 3C We can rename 3C as a new constant, let's call it C': y^3 = -(3/4)e^(x) + C'Final solution: To solve for y, we take the cube root of both sides: y = ±cbrt(-(3/4)e^(x) + C') This matches answer choice (A), which is y = ±root(3)(-(3e^(x))/(4) + C).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the equation.

(dy)/(dx)=-(1)/(4)e^(x)y^(-2)
Choose 1 answer:
(A)
y=+-root(3)(-(3e^(x))/(4)+C)
(B)
y=root(3)(-(3e^(x))/(4)+C)
(C)
y=+-root(3)(-(3e^(x))/(4))+C
(D)
y=root(3)(-(3e^(x))/(4))+C

Solve the equation. $\newline$ $\frac{d y}{d x}=-\frac{1}{4} e^{x} y^{-2}$ $\newline$ Choose $1$ answer: $\newline$ (A) $y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}+C}$ $\newline$ (B) $y=\sqrt[3]{-\frac{3 e^{x}}{4}+C}$ $\newline$ (C) $y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}}+C$ $\newline$ (D) $y=\sqrt[3]{-\frac{3 e^{x}}{4}}+C$

View step-by-step help

Home

Math Problems

Calculus

Find derivatives of using multiple formulae

Full solution

Q. Solve the equation.

\newline

\frac{d y}{d x}=-\frac{1}{4} e^{x} y^{-2}

\newline

Choose

1

answer:

\newline

(A)

y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}+C}

\newline

(B)

y=\sqrt[3]{-\frac{3 e^{x}}{4}+C}

\newline

(C)

y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}}+C

\newline

(D)

y=\sqrt[3]{-\frac{3 e^{x}}{4}}+C

Separate variables: We are given the differential equation $(\frac{dy}{dx} = -\frac{1}{4}e^{x}y^{-2})$ . This is a separable differential equation, which means we can separate the variables $y$ and $x$ on different sides of the equation.
Multiply and integrate: To separate the variables, we multiply both sides by $y^2$ and $dx$ to get $y^2 dy = -\frac{1}{4}e^{x} dx$ .
Integrate both sides: Now we integrate both sides of the equation. The left side with respect to $y$ , and the right side with respect to $x$ . $\int y^2 \, dy = \int -(\frac{1}{4})e^{x} \, dx$
Simplify and rename constant: The integral of $y^2$ with respect to $y$ is $(1/3)y^3$ . The integral of $-(1/4)e^{x}$ with respect to $x$ is $-(1/4)e^{x}$ . So we have: $\newline$ $(1/3)y^3 = -(1/4)e^{x} + C$ , where $C$ is the constant of integration.
Take cube root: We multiply through by $3$ to get rid of the fraction on the left side: $\newline$ $y^3 = -\left(\frac{3}{4}\right)e^{x} + 3C$ $\newline$ We can rename $3C$ as a new constant, let's call it $C'$ : $\newline$ $y^3 = -\left(\frac{3}{4}\right)e^{x} + C'$
Final solution: To solve for $y$ , we take the cube root of both sides: $\newline$ $y = \pm\sqrt[3]{-(\frac{3}{4})e^{x} + C'}$ $\newline$ This matches answer choice (A), which is $y = \pm\sqrt[3]{-(\frac{3e^{x}}{4}) + C}$ .