Question

Use separation of variables to solve the differential equation with initial condition y(1)=-2​

Answer 1

Answer: A

Step-by-step explanation:

$\frac{dy}{dx}=\frac{1+x}{xy} \\ \\ y \text{ } dy=\frac{1+x}{x} \text{ } dx \\ \\ y \text{ } dy= \left(\frac{1}{x}+1 \right) \text{ } dx \\ \\ \int y \text{ } dy=\int \left(\frac{1}{x}+1 \right) \text{ } dx \\ \\ \frac{y^{2}}{2}=\ln \left| x \right|+x+C$

Substituting in the initial condition,

$\frac{(-2)^{2}}{2}=\ln \left|1 \right|+1+C\\\\2=1+C\\\\C=1$

So,

$\boxed{\frac{1}{2}y^{2}=\ln \left|x \right|+x+1}$

Answer 2

Answer:

$\textsf{A.} \quad \dfrac{1}{2}y^2=\ln|x|+x+1$

Step-by-step explanation:

To solve the differential equation:

Rearrange the given equation to get all the terms containing y on the left side, and all the terms containing x on the right side:

         $\dfrac{dy}{dx}=\dfrac{1+x}{xy}$

$\implies y\dfrac{dy}{dx}=\dfrac{1+x}{x}$

$\implies y\:dy=\dfrac{1+x}{x}\:dx$

$\implies y\:dy=\dfrac{1}{x}+\dfrac{x}{x}\:dx$

$\implies y\:dy=\left(\dfrac{1}{x}+1\right)\:dx$

Integrate both sides:

$\displaystyle \implies \int y\:dy=\int \left(\dfrac{1}{x}+1\right)\:dx$

$\implies \dfrac{1}{2}y^2=\ln|x|+x+C$

Find the value of C using the given values of x and y:

when x = 1, y = -2

$\implies \dfrac{1}{2}(-2)^2=\ln|1|+1+C$

$\implies 2=0+1+C$

$\implies C=1$

Substitute the found value of C:

$\implies \dfrac{1}{2}y^2=\ln|x|+x+1$