Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the general solution of the differential equation $(1+\text{y}^2)+(\text{x}-\text{e}^{{\tan^{-1}\text{y}}})\frac{\text{dy}}{\text{dx}}=0.$
✓

Answer

Given, differential equation is
$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow(1+\text{y}^2)\frac{\text{dy}}{\text{dx}}+\text{x}-\text{e}^{\tan^{-1}\text{y}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
This is a linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q},$ we get
$\text{P}=\frac{1}{1+\text{y}^2},\text{Q}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\text{I.F.}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dx}}$
$=\text{e}^{\tan^{-1}\text{y}}$
So, the general solution is,
$\text{x}.\text{e}^{\tan^{-1}\text{y}}=\int\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}.\text{e}^{\tan^{-1}\text{y}}\text{dx}+\text{C}$
Put $\text{e}^{\tan^{-1}\text{y}}=\text{t}$
$\Rightarrow\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy}=\text{dt}$
$\therefore\text{x}.\text{e}^{\tan^{-1}\text{y}}=\int\text{t}\text{dt}+\text{C}$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\frac{\text{t}^2}{2}+\text{C}$

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}$
→
A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of Rs. 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.
→
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $2\ m$ and volume is $8\ m^3.$ If building of tank costs $ Rs. 70$ per sq metres for the base and $Rs. 45$ per square metre for sides. What is the cost of least expensive tank?
→
Solve the differential equation $\text{dy}=\cos\text{x}(2-\text{y}\text{cosesx})\text{dx}$ given that $\text{y}=2$ when $\text{x}=\frac{\pi}{2}.$
→
If the sum of the length of the hypotenuse and a side of a right-angles triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{\pi}{3}.$
→
Prove that the semi $-$ vertical angle of the right circular cone of given volume and least curved surface area is $\cot ^{-1} \sqrt{2}$
→
Prove that:
$\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}=2(\text{a}+\text{b}+\text{c})^3$
→
Let N denote the set of all natural numbers and R be the relation on$\text{N} \times \text{N}$defined by $\text{(a, b) R (c, d)}$ if ad$\text{(b + c) = bc(a + d)}$. Show that R is an equivalence relation.
→
A metal box with a square base and vertical sides is to contain $1024\ cm^3.$ The material for the top and bottom costs $₹\ 5\ per\ cm^2 $ and the material for the sides costs $₹\ 2.50\ per\ cm^2.$ Find the least cost of the box.
→
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sin^{-1}\Bigg(\frac{{1}-\text{x}^{2}}{1+\text{x}^{2}}\Bigg), 0 <\text{x}<1$
→