Question

Determine the solution of the differential equation:

$\frac{dy}{dx} = \frac{y}{1+x}$

• A. $y=C(1+x)$
• B. $y=C(1+x)^2$
• C. $y=Ce^{(1+x)}$
• D. $y=Cx^2$

Answer 1

Answer: (A)

$y=C(1+x)$

Answer 2

The given differential equation is: $\frac{dy}{dx} = \frac{y}{1+x}$

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

Step 1: Separate the variables

Rearrange the equation to group terms involving $y$ on one side and terms involving $x$ on the other side: $\frac{dy}{y} = \frac{dx}{1+x}$

Step 2: Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $\int \frac{1}{y} dy = \int \frac{1}{1+x} dx$

Step 3: Perform the integration

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln|y|$. The integral of $\frac{1}{1+x}$ with respect to $x$ is $\ln|1+x|$. Remember to add a constant of integration, say $K$, on one side: $\ln|y| = \ln|1+x| + K$

Step 4: Solve for y

To eliminate the logarithms, express the constant $K$ as $\ln|C'|$ (where $C'$ is a positive constant) or simply use the properties of logarithms. $\ln|y| - \ln|1+x| = K$ Using the logarithm property $\ln a - \ln b = \ln(a/b)$: $\ln\left|\frac{y}{1+x}\right| = K$ Now, exponentiate both sides: $e^{\ln\left|\frac{y}{1+x}\right|} = e^K$ $\left|\frac{y}{1+x}\right| = e^K$ Let $e^K = A$, where $A$ is a positive constant. $\frac{y}{1+x} = \pm A$ Let $C = \pm A$. Since $A$ is a positive constant, $C$ can be any non-zero real number. Also, note that $y=0$ is a solution to the original differential equation (since $\frac{d(0)}{dx} = 0$ and $\frac{0}{1+x} = 0$). If we set $C=0$ in our solution, we get $y=0$. Thus, $C$ can be any real number. $\frac{y}{1+x} = C$ Finally, solve for $y$: $y = C(1+x)$