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

Use separation of variables to solve the differential equation with initial condition-example-1
User AlmasB
by
6.6k points

2 Answers

9 votes

Answer: A

Explanation:


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

Substituting in the initial condition,


((-2)^(2))/(2)=\ln \left|1 \right|+1+C\\\\2=1+C\\\\C=1

So,


\boxedx \right

User Roland Illig
by
7.2k points
8 votes

Answer:


\textsf{A.} \quad (1)/(2)y^2=\ln|x|+x+1

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:


(dy)/(dx)=(1+x)/(xy)


\implies y(dy)/(dx)=(1+x)/(x)


\implies y\:dy=(1+x)/(x)\:dx


\implies y\:dy=(1)/(x)+(x)/(x)\:dx


\implies y\:dy=\left((1)/(x)+1\right)\:dx

Integrate both sides:


\displaystyle \implies \int y\:dy=\int \left((1)/(x)+1\right)\:dx


\implies (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 (1)/(2)(-2)^2=\ln|1|+1+C


\implies 2=0+1+C


\implies C=1

Substitute the found value of C:


\implies (1)/(2)y^2=\ln|x|+x+1

User Fyngyrz
by
7.2k points
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