Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
✓

Answer

Here, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\Big\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\Big\}$ 
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}\Big\{\log\Big(\frac{\text{vx}}{\text{x}}\Big)+1\Big\}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v + v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v}$
$\int\frac{1}{\text{v}\log\text{v}}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\log\log\text{v}=\log|\text{x}|+\log\text{C}$
$\log\text{v}=\text{xC}$
$\log\frac{\text{y}}{\text{x}}=\text{xC}$
$\frac{\text{y}}{\text{x}}=\text{e}^{\text{xC}}$
$\text{y}=\text{xe}^{\text{xC}}$

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 "4 Marks" from DIFFERENTIAL EQUATIONS→DIFFERENTIAL EQUATIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The probabilities of two students A and B coming to the school in time are $\frac{3}{7}\text{ and }\frac{5}{7}$ respectively. Assuming that the events, ‘A coming in time’ and ‘B coming in time’ are independent, find the probability of only one of them coming to the school in time. Write at least one advantage of coming to school in time.
→
Show that four points whose position vectors are
$6\hat{\text{i}}-7\hat{\text{j}},16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},3\hat{\text{i}}-6\hat{\text{k}},2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
→
If X follows a binomial distribution with mean 4 and variance 2, find P (X ≥ 5).
→
Prove that:
$\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
→
Find the distance of the point (- 1, - 5, - 10), from the point of intersection of the line
$\overrightarrow{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k})}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k})}\text{ and the plane}\overrightarrow{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k})}=5.$
→
If $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$, find A-1. Using A-1, solve the system of linear equations:
x - 2y = 10, 2x + y + 3z = 8, -2y + z = 7
→
Find the angle between the follwing pairs of lines:

$\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$ and $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}-\mu\big(2\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}\big)$

→
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
→
The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}.$ Find the probability that among 5 components tested.
  1. exactly 2 will survive.
  2. at most 3 will survive.
→
Find the equatoion of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.
→