Question 30 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises 1 to 10, find the general solution:

dydx=(1+x2)(1+y2)\frac{dy}{dx} = (1 + x^2) (1 + y^2)
Correct Answer

The given differential equation is:

dydx=(1+x2)(1+y2)\frac{dy}{dx} = (1 + x^2)(1 + y^2)

Step 1: Separate the variables

Rearranging the equation to separate variables xx and yy:

dy1+y2=(1+x2) dx\frac{dy}{1 + y^2} = (1 + x^2)\, dx

Step 2: Integrate both sides

Integrating both sides:

∫dy1+y2=∫(1+x2) dx\int \frac{dy}{1 + y^2} = \int (1 + x^2)\, dx

Step 3: Apply standard integrals

Using the standard integrals:

∫dy1+y2=tan⁡−1(y)\int \frac{dy}{1 + y^2} = \tan^{-1}(y) ∫(1+x2) dx=x+x33\int (1 + x^2)\, dx = x + \frac{x^3}{3}

Step 4: Combine and simplify

Therefore:

tan⁡−1(y)=x+x33+C\tan^{-1}(y) = x + \frac{x^3}{3} + C

Where CC is the constant of integration.

Step 5: Express y in terms of x

Taking tangent on both sides:

y=tan⁡ ⁣(x+x33+C)y = \tan\!\left(x + \frac{x^3}{3} + C\right)

This is the general solution of the given differential equation.

Exercise: EXERCISE 9.3 | Q: 6 | (Page 12)
For More Understanding

Explanation

This is a separable differential equation where variables can be isolated on opposite sides. The key integration formulas used are:

∫dy1+y2=tan⁡−1(y)\int \frac{dy}{1+y^2} = \tan^{-1}(y) ∫(1+x2) dx=x+x33\int (1+x^2)\, dx = x + \frac{x^3}{3}

After integration, we express yy explicitly in terms of xx to obtain the general solution with arbitrary constant CC.

Solution Steps

  1. Separate variables:

    dy1+y2=(1+x2) dx\frac{dy}{1+y^2} = (1+x^2)\, dx

  2. Integrate both sides:

    ∫dy1+y2=∫(1+x2) dx\int \frac{dy}{1+y^2} = \int (1+x^2)\, dx

  3. Apply standard integrals:

    tan⁡−1(y)=x+x33+C\tan^{-1}(y) = x + \frac{x^3}{3} + C

  4. Express y:

    y=tan⁡ ⁣(x+x33+C)y = \tan\!\left(x + \frac{x^3}{3} + C\right)
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