Question

Find the solution to the differential equation:

$\frac{dy}{dx} = y \cdot sin(x)$

Answer 1

$y = K e^{-\cos(x)}$

Answer 2

The given differential equation is: $\frac{dy}{dx} = y \cdot \sin(x)$

This is a first-order ordinary differential equation that can be solved using the method of separation of variables.

Step 1: Separate the variables.

Divide both sides by $y$ (assuming $y \neq 0$) and multiply both sides by $dx$: $\frac{dy}{y} = \sin(x) dx$

Step 2: Integrate both sides. $\int \frac{dy}{y} = \int \sin(x) dx$

Step 3: Evaluate the integrals.

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln|y|$. The integral of $\sin(x)$ with respect to $x$ is $-\cos(x)$. Remember to add a constant of integration, $C$, on one side. $\ln|y| = -\cos(x) + C$

Step 4: Solve for $y$.

To eliminate the natural logarithm, exponentiate both sides with base $e$: $e^{\ln|y|} = e^{-\cos(x) + C}$ $|y| = e^{-\cos(x)} \cdot e^C$ Let $e^C = A$, where $A$ is a positive constant ($A > 0$). $|y| = A \cdot e^{-\cos(x)}$ This implies $y = \pm A \cdot e^{-\cos(x)}$. Let $K = \pm A$. Since $A$ is a positive constant, $K$ can be any non-zero real constant. $y = K \cdot e^{-\cos(x)}$

Step 5: Consider the trivial solution.

If $y=0$, then $\frac{dy}{dx} = 0$. Substituting $y=0$ into the original differential equation gives $0 = 0 \cdot \sin(x)$, which is $0=0$. So, $y=0$ is also a solution. Our general solution $y = K \cdot e^{-\cos(x)}$ includes the trivial solution $y=0$ if we allow $K=0$. Therefore, $K$ can be any real constant.

The general solution to the differential equation is: $y = K e^{-\cos(x)}$ where $K$ is an arbitrary real constant.

Explanation of the solution:

The given differential equation is a first-order separable ordinary differential equation. Separate the variables $y$ and $x$ to opposite sides of the equation. Integrate both sides. The integral of $1/y$ is $\ln|y|$, and the integral of $\sin(x)$ is $-\cos(x)$. Introduce an integration constant $C$. Exponentiate both sides to solve for $y$. Let $e^C = A$ (a positive constant) and then $y = \pm A e^{-\cos(x)}$. Combine $\pm A$ into a single arbitrary constant $K$, which can be any real number (including zero, which accounts for the $y=0$ trivial solution).