Question

If the general solution of $(1+y^2)dx = (\tan^{-1}y - x)dy$ is $x = f(y)+ce^{-\tan^{-1}y}$, then $f(y)=$

• A. $\tan^{-1}y$
• B. $\tan^{-1}y+1$
• C. $\tan^{-1}y-1$
• D. $y\tan^{-1}y$

Hint:

To solve a first-order linear differential equation of the form $y' + P(x)y = Q(x)$, first calculate the integrating factor $I(x) = e^{\int P(x)dx}$. The solution is then given by $y \cdot I(x) = \int Q(x)I(x)dx + C$. Remember that the variables can be swapped, as in this problem where $x$ is a function of $y$.

Answer 1

The Correct Option is C

Step 1: Rearrange the differential equation into standard linear form.
The given equation is $(1+y^2)dx = (\tan^{-1}y - x)dy$. Divide by $dy$ and rearrange to treat $x$ as the dependent variable and $y$ as the independent variable. \[ (1+y^2)\frac{dx}{dy} = \tan^{-1}y - x. \] \[ \frac{dx}{dy} = \frac{\tan^{-1}y}{1+y^2} - \frac{x}{1+y^2}. \] \[ \frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2}. \] This is a linear first-order differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$.

Step 2: Identify $P(y)$ and $Q(y)$ and find the integrating factor (I.F.).
$P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan^{-1}y}{1+y^2}$. The integrating factor is $I.F. = e^{\int P(y) dy}$. \[ I.F. = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1}y}. \]

Step 3: Write down the general solution using the integrating factor.
The solution to a linear DE is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$. \[ x \cdot e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2} e^{\tan^{-1}y} dy + C. \]

Step 4: Evaluate the integral on the right-hand side.
Let $t = \tan^{-1}y$. Then $dt = \frac{1}{1+y^2} dy$. The integral becomes $\int t e^t dt$. We use integration by parts: $\int u dv = uv - \int v du$. Let $u=t$ and $dv=e^t dt$. Then $du=dt$ and $v=e^t$. \[ \int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1). \] Substituting back $t=\tan^{-1}y$, the integral is $e^{\tan^{-1}y}(\tan^{-1}y - 1)$.

Step 5: Find the final solution for x and identify f(y).
The solution is: \[ x \cdot e^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C. \] Divide by $e^{\tan^{-1}y}$: \[ x = (\tan^{-1}y - 1) + C e^{-\tan^{-1}y}. \] Comparing this to the given form $x = f(y)+ce^{-\tan^{-1}y}$, we can identify $f(y)$. \[ f(y) = \tan^{-1}y - 1. \]