Question

Solve the initial value problem. dy/dx=2xy, y(0)=3.

Answer 1

The initial value problem dy/dx = 2xy, y(0) = 3 is solved by separating variables, integrating both sides, finding the integration constant using the initial condition, and then expressing y explicitly. The final solution is y = 3e^(x^2).

Step-by-step explanation:

We need to solve the initial value problem given by the differential equation dy/dx = 2xy, with the initial condition y(0) = 3. This is a separable differential equation, so we can separate variables and integrate both sides. The solution involves integrating the derivative of y with respect to x and using integration techniques to solve it.

1. Separate the variables: dy/y = 2x dx.
2. Integrate both sides: ∫ dy/y = ∫ 2x dx which gives ln|y| = x² + C.
3. Using the initial condition, we find C: ln|3| = 0² + C, thus C = ln(3).
4. Now we have the particular solution: ln|y| = x² + ln(3).
5. Exponentiate both sides to solve for y: y = e² + ln(3) or y = 3e².

The final solution to the initial value problem is y = 3e².

Answer 2

The solution to the initial value problem
`\( (dy)/(dx) = 2xy \)`, with the initial condition
`\( y(0) = 3 \), is \( y(x) = 3e^(x^2) \).`

Step-by-step explanation:

To solve the given initial value problem, we can use separation of variables. Start by separating variables and integrating both sides:

`\[ (1)/(y) \, dy = 2x \, dx \]`

Integrate both sides:

`\[ \ln|y| = x^2 + C \]`

Now, apply the initial condition
`\( y(0) = 3 \)`:

`\[ \ln|3| = 0 + C \]`

`\[ C = \ln|3| \]`

Substitute
`\( C \)` back into the equation:

`\[ \ln|y| = x^2 + \ln|3| \]`

Exponentiate both sides:

`\[ |y| = 3e^(x^2) \]`

Consider the absolute value, and since
`\( y(0) = 3 \)` is positive:

`\[ y(x) = 3e^(x^2) \]`

This is the solution to the initial value problem. The exponential term
`\( e^(x^2) \)` reflects the growth of the solution, and the initial condition ensures that the constant
`\( C \)` is determined, resulting in the specific solution
`\( y(x) = 3e^(x^2) \).`