Question Statement

Solve the differential equation:

$\frac{dy}{dx}+\left(\frac{1+x}{x}\right)y=0$

subject to the initial condition $y(1)=1$.

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Background and Explanation

This is a first-order separable ordinary differential equation. The method of separation of variables is used: rearrange so all $y$-terms are on one side and all $x$-terms on the other, integrate both sides, then apply the initial condition to determine the constant of integration and obtain the particular solution.

Relevant SLOs: Separable variable equations (M-12-A-82), Initial Value Problems (M-12-A-81).

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Solution

Step 1 — Separate the Variables

Rearrange the equation:

$\frac{dy}{dx}=-\left(\frac{1+x}{x}\right)y$

Divide both sides by $y$ and multiply by $dx$:

$\frac{dy}{y}=-\left(\frac{1+x}{x}\right)dx$

Step 2 — Decompose the Fraction

Split the right-hand side:

$\frac{1+x}{x}=\frac{1}{x}+1$

So:

$\frac{dy}{y}=-\left(\frac{1}{x}+1\right)dx$

Step 3 — Integrate Both Sides

$\int \frac{dy}{y}=-\int \left(\frac{1}{x}+1\right)dx$

$\ln y=-\ln x-x+c$

Step 4 — Combine Logarithms

Using $\ln a+\ln b=\ln (ab)$:

$\ln y+\ln x=-x+c$

$\ln (xy)=-x+c$

Step 5 — Apply the Initial Condition

Substitute $x=1,y=1$ into equation (2):

$\ln (1\cdot 1)=-1+c$

$0=-1+c⟹c=1$

Step 6 — State the Particular Solution

Substitute $c=1$ back into equation (2):

$\boxed{\ln (xy)=1-x}$

This can also be written as:

$\ln (xy)+x=1$

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Key Formulas and Methods

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Summary of Steps

1. Separate variables: $\frac{dy}{y}=-\left(\frac{1}{x}+1\right)dx$
2. Integrate both sides: $\ln y=-\ln x-x+c$
3. Combine logarithms: $\ln (xy)=-x+c$
4. Apply initial condition $y(1)=1$: find $c=1$
5. Particular solution: $\ln (xy)=1-x$