Solution of (x + y - 1)dx + (2x + 2y - 3)dy = 0 is

Download App

MCQ

Solution of $(x + y - 1)dx + (2x + 2y - 3)dy = 0$ is

A
$y + x + \log (x + y - 2) = c$
B
$y + 2x + \log (x + y - 2) = c$
✓
$2y + x + \log (x + y - 2) = c$
D
$2y + 2x + \log (x + y - 2) = c$

✓

Answer

Correct option: C.

$2y + x + \log (x + y - 2) = c$

c
(c) Given equation is $\frac{{dy}}{{dx}} = - \left( {\frac{{x + y - 1}}{{2x + 2y - 3}}} \right)$

Put $x + y = t$ ==> $\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} - 1$

$\therefore \frac{{dy}}{{dx}} = \frac{{1 - t}}{{2t - 3}}$ ==> $\frac{{dt}}{{dx}} - 1 = \frac{{1 - t}}{{2t - 3}}$ ==> $\frac{{dt}}{{dx}} = \frac{{t - 2}}{{2t - 3}}$

==> $\frac{{2t - 3}}{{t - 2}}dt = dx$. Integrating both sides, we get
$\int_{}^{} {\frac{{2t - 4}}{{t - 2}}dt} - \int_{}^{} {\frac{{3 - 4}}{{t - 2}}dt} = \int_{}^{} 1 dx$

==> $2t + \log (t - 2) = x + c$

==> $2(x + y) + \log (x + y - 2) = x + c$

==> $2y + x + \log (x + y - 2) = c$.

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 "M.C.Q (1 Marks)" from 9. differential equations→9. 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

There are $3$ bags, each containing $5$ white balls and $3$ black balls. Also there are $2$ bags, each containing $2$ white balls and $4$ black balls. A white ball is draws at rondom. Find the probability that this white ball is from a bag of the first group.

→

If $u = \int_{}^{} {{e^{ax}}\cos bx\;dx} $ and $v = \int_{}^{} {{e^{ax}}\sin bx\;dx} $, then $({a^2} + {b^2})({u^2} + {v^2}) = $

→

The general solution of differention eqution of the type $\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$ is:

$\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$
$\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
$\text{xe}^{\int\text{P}_{1}\text{dx}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$

→

$\int_0^{\pi /2} {\frac{{\sqrt {\cot x} }}{{\sqrt {\cot x} + \sqrt {\tan x} }}\,dx = } $

→

The period of $f(x) = x - [x]$, if it is periodic, is

→

The mid-point of the domain of the function $f(x)=\sqrt{4-\sqrt{2 x+5}}$ real $x$ is

→

The area bounded by the curve $y = e^x$ and the lines $y = \left| {x - 1} \right|,x = 2$ is given by

→

$\int_{}^{} {{{\tan }^{ - 1}}x\,dx = } $

→

Area lying between the curves y² = 4x and y = 2x is: