How do you find the particular solution to #2xy'-lnx^2=0# that satisfies y(1)=2?

Question

Scarlett Allison · Accepted Answer

Use the separation of variables method.
Integrate.
Use #y(1) = 2# to solve for the value of the integration constant.

Separate variables: #dy = ln(x^2)/(2x)dx# Integrate: #y = intln(x^2)/(2x)dx# #y = intln(x)/(x)dx# #y = ln^2(x) + C# Use #y(1) = 2# to solve for c: #2 = ln^2(1) + C# #2 = C# The particular solution is: #y = ln^2(x) + 2#

Emilia Aguilar · Answer

undefined To find the particular solution to the differential equation $2xy' - \ln(x^2) = 0$ that satisfies the initial condition $y(1) = 2$, you can follow these steps:

1. First, rewrite the equation in the form $y' = f(x,y)$ to separate variables.

2. Solve the resulting first-order ordinary differential equation for $y$ in terms of $x$.

3. Integrate both sides of the equation to find the general solution, including a constant of integration.

4. Apply the initial condition $y(1) = 2$ to determine the value of the constant of integration and obtain the particular solution.

By following these steps, you can find the particular solution that satisfies the given initial condition.

HIX Tutor · Answer

undefined To find the particular solution to the equation $2xy' - \ln(x^2) = 0$ that satisfies $y(1) = 2$, follow these steps:

1. **Rewrite the equation**: Start with the equation:
   $$
   2xy' = \ln(x^2)
   $$

2. **Simplify $\ln(x^2)$**: Remember that $\ln(x^2) = 2\ln(x)$. So, it becomes:
   $$
   2xy' = 2\ln(x)
   $$

3. **Divide both sides by 2**: This simplifies down to:
   $$
   xy' = \ln(x)
   $$

4. **Separate variables**: Rewrite $y'$ as $\frac{dy}{dx}$:
   $$
   y' = \frac{\ln(x)}{x}
   $$

5. **Integrate both sides**: We need to integrate:
   $$
   \int y' \, dx = \int \frac{\ln(x)}{x} \, dx
   $$

6. **Use integration by parts for $\int \frac{\ln(x)}{x} \, dx$**:
   - Let $u = \ln(x)$ and $dv = \frac{1}{x}dx$.
   - Then $du = \frac{1}{x}dx$ and $v = x$.
   
   So,
   $$
   \int \ln(x) \cdot \frac{1}{x} dx = x \ln(x) - \int x \cdot \frac{1}{x} dx = x \ln(x) - x + C
   $$

7. **Combine the results**: We get:
   $$
   y = x \ln(x) - x + C
   $$

8. **Use the initial condition $y(1) = 2$** to find $C$:
   $$
   y(1) = 1 \cdot \ln(1) - 1 + C = 0 - 1 + C = C - 1
   $$
   Set this equal to 2:
   $$
   C - 1 = 2 \implies C = 3
   $$

9. **Write the particular solution**: Finally, substitute $C$ back into the equation:
   $$
   y = x \ln(x) - x + 3
   $$

Thus, the particular solution is:
$$
y = x \ln(x) - x + 3
$$

HIX Tutor · Answer

undefined To find the particular solution for the equation $2xy' - \ln(x^2) = 0$ with the condition $y(1) = 2$, follow these steps:

1. **Rewrite the equation**: Start with the equation:
   $$
   2xy' = \ln(x^2)
   $$
   This means:
   $$
   y' = \frac{\ln(x^2)}{2x}
   $$

2. **Simplify $\ln(x^2)$**: Remember that $\ln(x^2) = 2\ln(x)$:
   $$
   y' = \frac{2\ln(x)}{2x} = \frac{\ln(x)}{x}
   $$

3. **Integrate**: To find $y$, integrate $y'$:
   $$
   y = \int \frac{\ln(x)}{x} \, dx
   $$

4. **Use integration by parts**: Let $u = \ln(x)$ and $dv = \frac{1}{x}dx$. Then $du = \frac{1}{x}dx$ and $v = x$. Now apply the integration by parts formula $ \int u \, dv = uv - \int v \, du $:
   $$
   y = x \ln(x) - \int x \cdot \frac{1}{x} \, dx
   $$
   $$
   y = x \ln(x) - \int 1 \, dx
   $$
   $$
   y = x \ln(x) - x + C 
   $$
   Where $C$ is the constant of integration.

5. **Use the initial condition $y(1) = 2$**: Plug in $x = 1$ and $y = 2$:
   $$
   2 = 1 \cdot \ln(1) - 1 + C
   $$
   Since $\ln(1) = 0$, this simplifies to:
   $$
   2 = -1 + C
   $$
   Adding 1 to both sides:
   $$
   C = 3
   $$

6. **Write the particular solution**: Substitute $C$ back into the equation:
   $$
   y = x \ln(x) - x + 3
   $$

So, the particular solution is:
$$
y = x \ln(x) - x + 3
$$