Download App

DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\frac{\text{y}}{\text{x}}=0,\text{y}(1)=0$
✓

Answer

$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\frac{\text{y}}{\text{x}}=0,\text{y}(1)=0$
This is a homogeneous equation, Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}-\text{v + cosec v}=0$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{cosec v}$
$\frac{\text{dv}}{\text{cosec v}}=\frac{\text{dx}}{\text{x}}$
$\sin\text{v dv}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\int\sin\text{v dv}=\int\frac{\text{dx}}{\text{x}}$
$-\cos\text{v}=\log_{\text{e}}\text{x + C}$
$-\cos\text{v}+\log_{\text{e}}\text{x}=\text{C}$
$\cos\text{v}+\log_{\text{e}}\text{x}=-\text{C}$
$\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\log_{\text{e}}\text{x}=-\text{C}$
As y(1) = 0
$\cos\Big(\frac{0}1\Big)=0+\log_{\text{e}}1=-\text{C}$
$1+0=-\text{C}$
$\Rightarrow\ \text{C}=-1$
$\Rightarrow\ \cos\Big(\frac{\text{y}}{\text{x}}\Big)+\log_{\text{e}}\text{x}=1$

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 $\text{A} = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 &1 & -2 \end{pmatrix}$ find $A^{–1}.$
Hence using $A^{–1}$ solve the system of equations $2x – 3y + 5z = 11, 3x + 2y – 4z = – 5, x + y – 2z = – 3.$
→
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\text{x}}{2}-\sin\frac{\pi\text{x}}{6}\text{ on }[-1,0]$
→
Integrate the function in Exercise:$\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}$
→
Find the equation of the plane passing through the line of intersection of the planes $x + 2y +3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y - 6z + 8 = 0.$
→
Find the shortest distance between the lines
$\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}$
→
Find the value of 'a' for which the function f defined by
 $\text{f}\text{(x)}=\begin{cases}\text{a}\sin\frac{\pi}{2}(\text{x}+1),& \text{x}\leq0 \\\frac{\tan\text{x-sin}\text{x}}{\text{x}^3} &\text{x} > 0\end{cases}$ is discontinuous at x = 0.
→
Prove that the points $\hat{\text{i}}-\hat{\text{j}},\ 4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ are the vertices of a right-angled triangle.
→
Prove that $|A \ adj \ A| = |A|^n.$
→
Prove that the curves $xy = 4$ and $x^2 + y^2 = 8$ touch each other.
→
Given $A=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$, find $B A$ and use this to solve the system of equations $y$ $+2 z=7, x-y=3,2 x+3 y+4 z=17$
→