Find the general solution of the differential equation
\[
\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}.
\]
Show Hint
Solution and Explanation
Step 1: Separate variables.
We can separate the variables \( x \) and \( y \) to integrate both sides:
\[
\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}.
\]
Step 2: Integrate both sides.
The integral of \( \frac{1}{1 + y^2} \) is \( \tan^{-1}(y) \), and the integral of \( \frac{1}{1 + x^2} \) is \( \tan^{-1}(x) \), so we get:
\[
\tan^{-1}(y) = \tan^{-1}(x) + C,
\]
where \( C \) is the constant of integration.
Step 3: Solve for \( y \).
Thus, the general solution is:
\[
y = \tan(\tan^{-1}(x) + C).
\]
Step 4: Conclusion.
The general solution to the differential equation is:
\[
y = \tan(\tan^{-1}(x) + C).
\]
Top MPBSE Class XII Board Higher Mathematics Questions
- In the function \( f : \mathbb{R} \rightarrow \mathbb{R} \), given by \( f(x) = 5x \):
- \( \tan^{-1} \sqrt{3} - \sec^{-1}(-2) \) is equal to:
If \( y = \sqrt{e^x} \), \( x > 0 \), then \( \frac{dy}{dx} = \underline{\hspace{2cm}} \)
- Rate of change of area of circle per second with respect to its radius \( r \) when \( r = 5 \, \text{cm} \) will be \(\underline{\hspace{2cm}}\)
- \( \int \left( \sin^{-1} x + \cos^{-1} x \right) dx\) = \(\underline{\hspace{2cm}}\)
Top MPBSE Class XII Board Differential Equations Questions
- The number of arbitrary constants in the particular solution of a differential equation of third order are \(\underline{\hspace{2cm}}\)
Order of differential equation \( xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0 \) is 2.
- Integrating factor of differential equation \( \frac{dy}{dx} - y = \cos x \) is \( e^{-x} \).
Find the general solution of the differential equation \[ x \frac{dy}{dx} + 2y = x^2 \,\,\, \text{where} \,\,\, (x \neq 0). \]
Top MPBSE Class XII Board Questions
- What is the Market Economy?
- What is positive economics analysis?
- What is profit?
- What is average product?
- Write three relationships between total utility and marginal utility.