Solve the following differential equation dy dx =x^5 ^ -1 (x^3) - Vidyadip

Question

Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^5\tan^{-1}(\text{x}^3)$

✓

Answer

$\frac{\text{dy}}{\text{dx}}=\text{x}^5\tan^{-1}(\text{x}^3)$$\text{dy}=\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$
$\int\text{dy}=\int\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$
put $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
So,
$\int\text{dy}=\frac{1}{3}\Big[\tan^{-1}\text{t}\int\text{t dt}=\int\Big(\frac{1}{1+\text{t}^2}\Big)\times\int\text{t dx}\Big)\Big]\text{dt}+\text{C}$
Using integration by parts
$\text{y}=\frac{1}{3}\Big[\frac{\text{t}^2}{2}+\tan^{-1}-\int\frac{\text{t}^2}{2(\text{t}^2+1)}\text{dt}\Big]+\text{C}$
$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(\frac{\text{t}^2}{\text{t}^2+1}\Big)\text{dt}+\text{C}$
$\text{y}=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(1-\frac{1}{\text{t}^2+1}\Big)\text{dt}+\text{C}$
$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\frac{1}{6}\tan^{-1}\text{t}+\text{C}$
$\text{y}=\frac{1}{6}(\text{t}^2+1)\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\text{C}$
$\text{y}=\frac{1}{6}[(\text{t}^2+1)\tan^{-1}\text{t}-\text{t}]+\text{C}$
So,
$\text{y}=\frac{1}{6}[(\text{x}^6+1)\tan^{-1}(\text{x}^3)-\text{x}^3]+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Differential Equations→Differential Equations — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\text{A}=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix},\text{ B}=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix},$ find matrix X such that 2A + 3X = 5B.

The scalar product of the vector $\hat{\text{l}}+\hat{\text{j}}+\hat{\text{k}}$ with a unit a vector along the sum of vectors $2\hat{\text{l}}+4\hat{\text{j}}-5\hat{\text{k}}\ \text{and}\ \lambda\hat{\text{l}}+2\hat{\text{j}}+3\hat{\text{k}}$ is equal to one. Find the value of $\lambda.$

Find the distance of the point (2, 12, 5) from the point of intersection of the line $\vec{\text{r}}=2\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0.$

Evaluate the following definite integrals:
$\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$

If $\text{X}=\begin{bmatrix}3&1&-1\\5&-2&-3\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}2&1&-1\\7&2&4\end{bmatrix},$ then find:

X + Y
2X - 3Y
A matrix Z such that X + Y + Z is a zero matrix.

Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}$

Find the foot of the perpendicular drawn from the point A(1, 0, 3) to the joint of the points B(4, 7, 1) and C(3, 5, 3).

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$, then find the value of $\frac{\text{f}^{2}\text{y}}{\text{dx}}\text{at}\theta = \frac{\pi}{6}.$

Two schools $P$ and $Q$ want to award their selected students on the values of Discipline, Politeness and Punctuality. The school $P$ wants to award $₹ x$ each $,₹ y$ each and $₹ z$ each for the three respective values to its $3, 2$ and $1$ students with a total award money of $₹ 1,000$. School $Q$ wants to spend $₹ 1,500$ to award its $4, 1$ and $3$ students on the respective values $($by giving the same award money for the three values as before$)$. If the total amount of awards for one prize one each value is $₹ 600,$ using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.