3.8 Q-19

Question Statement

Find the general solution of the differential equation:

$$\frac{d y}{d x} - x = x y^{2}$$

Also, find the particular solution if $y = 1$ when $x = 0$.

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Background and Explanation

This is a first-order nonlinear differential equation. The method to solve this involves separation of variables, where we attempt to isolate terms involving $y$ on one side and terms involving $x$ on the other. Once the variables are separated, we integrate both sides to obtain the general solution.

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Solution

1. Rewrite the equation:

   Start with the given equation:

$$\frac{d y}{d x} - x = x y^{2}$$

2. Isolate the derivative term:

   Move all terms involving $y$ to one side and the terms involving $x$ to the other:

$$\frac{d y}{d x} = x + x y^{2}$$

3. Separate the variables:

   Factor out $x$ from the right-hand side:

$$\frac{d y}{d x} = x (1 + y^{2})$$

Now, separate the variables so that all terms with $y$ are on one side and all terms with $x$ are on the other:

$$\frac{1}{1 + y^{2}} , d y = x , d x \tag{i}$$

4. Integrate both sides:

   Integrating both sides:

$$\int \frac{1}{1 + y^{2}} , d y = \int x , d x$$

The integral on the left is a standard integral:

$$\int \frac{1}{1 + y^{2}} , d y = \tan^{-1} y$$

The integral on the right is straightforward:

$$\int x , d x = \frac{x^{2}}{2}$$

So, after integrating:

$$\tan^{-1} y = \frac{x^{2}}{2} + C \tag{a}$$

5. Find the particular solution:

   To find the particular solution, use the initial condition $y = 1$ when $x = 0$. Substitute these values into equation (a):

$$\tan^{-1}(1) = \frac{0^{2}}{2} + C$$

We know that $\tan^{-1}(1) = \frac{\pi}{4}$, so:

$$\frac{\pi}{4} = C$$

6. Write the particular solution:

   Now that we know $C = \frac{\pi}{4}$, substitute this into the general solution:

$$\tan^{-1} y = \frac{x^{2}}{2} + \frac{\pi}{4}$$

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Key Formulas or Methods Used

• Separation of Variables: Rearranging the equation to isolate the terms with $y$ on one side and $x$ on the other.
• Standard Integrals: Recognizing the integral $\int \frac{1}{1 + y^{2}} , d y = \tan^{-1} y$ and integrating $x$.
• Initial Conditions: Using the given initial values to find the constant of integration.

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Summary of Steps

1. Start with the equation: $\frac{d y}{d x} - x = x y^{2}$.
2. Rearrange to isolate $\frac{d y}{d x}$: $\frac{d y}{d x} = x(1 + y^{2})$.
3. Separate the variables: $\frac{1}{1 + y^{2}} , d y = x , d x$.
4. Integrate both sides: $\tan^{-1} y = \frac{x^{2}}{2} + C$.
5. Use the initial condition $y = 1$ when $x = 0$ to find $C$, giving $C = \frac{\pi}{4}$.
6. The particular solution is: $\tan^{-1} y = \frac{x^{2}}{2} + \frac{\pi}{4}$.