y' - y = 1, ;y(0) = - 1 નો ઉકેલ y(x) = - Vidyadip

MCQ

$y' - y = 1,\;y(0) = - 1$ નો ઉકેલ $y(x) = $

A
$ - \exp (x)$
B
$ - \exp ( - x)$
✓
$-1$
D
$\exp (x) - 2$

✓

Answer

Correct option: C.

$-1$

(c) $\frac{{dy}}{{dx}} - y = 1$ ==> $\frac{{dy}}{{1 + y}} = dx$

Integrating both sides $\log (1 + y) = x + c$ ==> $1 + y = {e^x}.{e^c}$

$\because x=0,\,\,y=-1$.Then, $1 - 1 = e.{e^c}$ ==>${e^c} = 0$

Therefore solution $1 + y = {e^x} \times 0 \Rightarrow y(x) = - 1$.

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 "MCQ" from 9. differential equations→9. differential equations — all question types→ગણિત ધોરણ 12 સાયન્સ question papers→Gujarat Board ધોરણ 12 સાયન્સ question papers→Gujarat Board question paper generator→

Similar questions

$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $

$\left[ {\begin{array}{*{20}{c}}4&7\\1&2\end{array}} \right]$ નો વ્યસ્ત શ્રેણિક મેળવો.

ધારો કે $\mathrm{P}(\alpha, \beta, \gamma)$ એ બિંદુ $\mathrm{Q}(1,6,4)$ નું રેખા $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ પરનું પ્રતિબિંબ છે. તો $2 \alpha+\beta+\gamma=$ ...............

જો $f: N \rightarrow N , f( x )= x +3$ હોય, તો $f^{-1}( x )=\ .........$

જો $f(x)$ એ $x$ માં દ્રીઘાત બહુપદી છે તો $\int\limits_0^1 {f(x) dx}$ મેળવો.

જો રેખા $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ અને સમતલ $2x-y+\sqrt{\lambda}z+4={0}$ કે જેથી $\sin\theta=\frac{1}{3},$ તો $|3\lambda|=\ ........$

જો $p + q + r = 0 = a + b + c$, તો $\left| {\,\begin{array}{*{20}{c}}{pa}&{qb}&{rc}\\{qc}&{ra}&{pb}\\{rb}&{pc}&{qa}\end{array}\,} \right|= . . . $

$\int_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ ની કિમંત મેળવો.

ઊગમબિંદુથી રેખા $ \left\{ \left( 4+3k,2+4k,4-5k\right )|k \in R\right\} $ નું લંબઅંત૨ $...... .$

$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = $