Download App

Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following initial value problems:
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0,\text{ y}(0)=0$
✓

Answer

We have,
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0$
$\Rightarrow(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$
$\Rightarrow(1+\text{y}^2)\frac{\text{dx}}{\text{dy}}=-(\text{x}-\text{e}^{\tan^{-1}\text{y}})$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\frac{1}{1+\text{y}^2}$ and $\text{Q}=\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$
$=\text{e}^{\tan^{-1}\text{y}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{\tan^{-1}\text{y}},$ we get
$\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\Rightarrow\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\frac{1}{1+\text{y}^2}$
Integrating both sides with respect to y, we get
$\text{e}^{\tan^{-1}\text{y}}\text{x}=\int\frac{1}{1+\text{y}^2}\text{dy}+\text{C}$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}+\text{C}\ ...(2)$
Now,
$\text{y}(0)=0$
$\therefore\ 0\times\text{e}^{0}=0+\text{C}$
$\Rightarrow\text{C}=0$
Putting the value of C in (2), we get
$\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}+0$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}$
Hence, $\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}$ 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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3). Find the equation of the plane.
→
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 $\frac{\text{dy}}{\text{dx}}.$
→
A letter is known to have come either from $TATA \ NAGAR$ or from $CALCUTTA$. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from $TATA$ $NAGAR$.
→
Evaluate $\int\limits^{2}_{-1}(\text{e}^{3\text{x}} + 7\text{x} - 5) \text{dx}$ as a limits of sum.
→
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
→
Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
→
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dx}}{\text{dx}}=-\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
→
Discuss the continuity of the function f, where f is defined by: $\text{f(x)}= \begin{cases}\ 2\text{x},\ \ \text{if}\ \text{x}<0 \\0,\ \ \ \ \text{if}\ 0\leq\text{x}\leq1\\4\text{x},\ \ \ \text{if}\ \text{x}>1\end{cases}$
→
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 vector equation of the line passing through the point (1, -1, 2) and perpendicular to the plane 2x - y + 3z - 5 = 0.
→