Download App

DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation:
$\text{y e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\big(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}\big)\text{dy}$
✓

Answer

We have,
$\text{y e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\big(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}\big)\text{dy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}}{\text{y e}^{\frac{\text{x}}{\text{y}}}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{x}}{\text{y}}\text{e}^{\frac{\text{x}}{\text{y}}}+1}{\text{e}^{\frac{\text{x}}{\text{y}}}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}+\text{e}^{\frac{-\text{x}}{\text{y}}}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + y}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + y}\frac{\text{dv}}{\text{dx}}=\text{v + e}^{-\text{v}}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\text{e}^{-\text{v}}$
$\Rightarrow\ \text{e}^{\text{v}}\text{dv}=\frac{1}{\text{y}}\text{dy}$
Integrating both sides, we get
$\int\text{e}^{\text{v}}\text{dv}=\int\frac{1}{\text{y}}\text{dy}$
$\Rightarrow\ \text{e}^{\text{v}}=\log|\text{y}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \text{e}^{\frac{\text{x}}{\text{y}}}=\log|\text{y}|+\text{C}$
Hence, $\text{e}^{\frac{\text{x}}{\text{y}}}=\log|\text{y}|+\text{C}$ is the required solution.

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

In a certain college, $4\%$ of boys and $1\%$ of girls are taller than $1.75$ metres. Further more, $60\%$ of the students in the colleges are girls. A student selected at random from the college is found to be taller than $1.75$ metres. Find the probability that the selected students is girl.
→
Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
→
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$
→
Decompose the vector $6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ into vectors which are parallal and perpendicular to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
→
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:
  1. X + Y
  2. 2X - 3Y
  3. A matrix Z such that X + Y + Z is a zero matrix.
→
Differentiate w.r.t. x the function in Exercise:
$\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big],\ 0<\text{x}<\frac{\pi}{2}$
→
If $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$ and $\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix},$ verify that |AB| = |A| |B|.
→
If $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},$ find $A^2 - 5A - 14.$
→
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
→
Find the equation of a plane passing through the points $(0, 0, 0)$ and $(3, -1,2)$ and parallel to the line $\frac{\text{x}-4}{1}=\frac{\text{y}+3}{-4}=\frac{\text{z}+1}{7}$ 
→