Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\Rightarrow\text{dy}=(\tan^{-1}\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\tan^{-1}\text{x})\text{dx}$
$\Rightarrow\text{y}=\int1\times\tan^{-1}\text{x}\text{ dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x }-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$
So, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is the solution o the given differential equation.
$\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}$
$\Rightarrow\text{dy}=(\tan^{-1}\text{x})\text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int(\tan^{-1}\text{x})\text{dx}$
$\Rightarrow\text{y}=\int1\times\tan^{-1}\text{x}\text{ dx}$
$\Rightarrow\text{y}=\tan^{-1}\text{x}\int\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x }-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$
So, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\text{y}=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\log|1+\text{x}^2|+\text{C}$ is the solution o the given differential equation.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate the following integrals:$\int(\text{x}+1)\text{e}^{\text{x}}\log(\text{xe}^{\text{x}})\text{dx}$
→If $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
→If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$, then find a matrix $X$ such that $X A=B$.
→Evaluate the following intregals:
$\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
→$\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
If $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},$ find $A^2 - 5A - 14.$
→Evaluate the following integrals:$\int\frac{\text{x}+2}{2\text{x}^2+6\text{x}+5}\text{ dx}$
→Two tailors, A and B earn Rs. $15$ and Rs. $20$ per day respectively. A can stitch $6$ shirts and $4$ pants while $B$ can stitch $10$ shirts and $4$ pants per day. How many days shall each work if it is desired to produce (at least) $60$ shirts and $32$ pants at a minimum labour cost?
→We first draw the lines processed on three machines : Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by following table:
Formulate the above problem as L.P.P. Solve it graphically to get maximum profit.
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}$
→If $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big), 0<\text{x}<1$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
→