Solve dy/dx = y/x given y(1)=2

Question

Solve the differential equation $\dfrac{dy}{dx} = \dfrac{y}{x}$ , given the initial condition $y(1) = 2$ .

Solution — Step by Step

This is a separable differential equation — we can separate variables so that all $y$ terms are on one side and all $x$ terms are on the other.

Rewrite:

\frac{dy}{dx} = \frac{y}{x}

Separate variables:

\frac{dy}{y} = \frac{dx}{x}

\int \frac{dy}{y} = \int \frac{dx}{x}

\ln|y| = \ln|x| + C

where $C$ is the constant of integration.

\ln|y| - \ln|x| = C

\ln\left|\frac{y}{x}\right| = C

Exponentiating both sides:

\left|\frac{y}{x}\right| = e^C

Since $e^C$ is just a positive constant, let $A = \pm e^C$ (absorbing the absolute value), so:

\frac{y}{x} = A \implies y = Ax

This is the general solution: $y = Ax$ for some constant $A$ .

We’re told $y(1) = 2$ , meaning when $x = 1$ , $y = 2$ . Substitute:

2 = A \times 1 \implies A = 2

Particular solution:

\boxed{y = 2x}

Check: $dy/dx = 2$ . And $y/x = 2x/x = 2$ . ✓

Check initial condition: $y(1) = 2(1) = 2$ . ✓

Why This Works

The equation $dy/dx = y/x$ is saying: the rate of change of $y$ with respect to $x$ is proportional to the ratio $y/x$ . This type of relationship arises in problems where growth is proportional to current value relative to some baseline.

The separation of variables technique works whenever we can write the equation as $f(y)dy = g(x)dx$ . We then integrate both sides independently — each side depends on only one variable.

Alternative Method

We can also recognise this as a homogeneous differential equation (degree 0 in $x$ and $y$ ). Substitute $y = vx$ where $v = y/x$ :

$y = vx \implies dy/dx = v + x(dv/dx)$

$v + x\frac{dv}{dx} = v \implies x\frac{dv}{dx} = 0 \implies dv/dx = 0$

So $v = \text{constant}$ , i.e., $y/x = \text{constant}$ , confirming $y = Ax$ .

This method is overkill for this equation but is the standard approach for harder homogeneous equations.

Common Mistake

A common error is integrating to get $y^2/2 = x^2/2 + C$ — this happens when students integrate $y\,dy$ instead of $dy/y$ . The correct integral of $1/y$ with respect to $y$ is $\ln|y|$ , not $y^2/2$ . Always check whether you have $dy/y$ or $y\,dy$ before integrating.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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