The solution of differential equation dy dx + y = 1 is - Vidyadip

MCQ

The solution of differential equation $\frac{{dy}}{{dx}} + y = 1$ is

✓
$y = 1 + c\,{e^{ - x}}$
B
$y = 1 - c\,{e^{ - x}}$
C
$y = x + c\,{e^{ - x}}$
D
$y = x - c\,{e^{ - x}}$

✓

Answer

Correct option: A.

$y = 1 + c\,{e^{ - x}}$

a
(a) $\frac{{dy}}{{dx}} + y = 1$; $I.F. $$ = {e^{\int {Pdx} }} = {e^{\int {dx} }} = {e^x}$

Hence solution is $y.{e^x} = \int {{e^x}dx + c} $

$y{e^x} = {e^x} + c$ ==> $y = 1 + c{e^{ - x}}$.

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

Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:

2
0
1
-1

Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:

$6-3\sqrt{2}$
$6-\sqrt{2}$
$6+3\sqrt{2}$
$6+\sqrt{2}$

If $f(x) = \left\{ {\begin{array}{*{20}{c}}{{e^x} + ax,}&{x < 0}\\{b{{(x - 1)}^2},}&{x \ge 0}\end{array}} \right.$ is differentiable at $x = 0,$ then $(a,\,b)$ is

The following system of equations $3x - 7y + 5z = 3; 3x + y + 5z = 7$ and $2x + 3y + 5z = 5$ are

If $A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&2&{ - 3}\\2&1&0\end{array}} \right]$ and $B = (adj\,A)$, and $C = 5A,$ then $\frac{{|adjB|}}{{|C|}}$=

$\int_{\,0}^{\,1} {\,\sin \left( {2{{\tan }^{ - 1}}\sqrt {\frac{{1 + x}}{{1 - x}}} } \right)\,dx = } $

Two dice are rolled one after the other. The probability that the number on the first is smaller than the number on the second is

Integrating factor of $\frac{{dy}}{{dx}} + \frac{y}{x} = {x^3} - 3$ is

$\mathop {{\rm{lim}}\,}\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

If $f(x)$ is a differentiable function such that $f:R \to r$ and $f\left( {\frac{1}{n}} \right) = 0\;\forall \;n \ge 1,n \in I$ then