Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
MCQ
Choose the correct answer from the given four options.The general solution of $\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$ is:
  • A
    $\text{e}^{\text{x}^{2}-\text{y}}=\text{C}$
  • B
    $\text{e}^{-\text{y}}+\text{e}^{\text{x}^{2}}=\text{C}$
  • ✓
    $\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
  • D
    $\text{e}^{\text{x}^{2}+\text{y}}=\text{C}$
✓

Answer

Correct option: C.
$\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
We have $\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$

$\Rightarrow\text{e}^{\text{y}}=\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}}$

$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=2\int\text{x}\text{e}^{\text{x}^{2}}\text{dx}$

Put $\text{x}^2=\text{t}$ in R.H.S. integral we get

$2\text{x}\text{dx}=\text{dt}$

$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int\text{e}^{\text{t}}\text{dt}$

$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{t}}+\text{C}$

$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}^2}+\text{C}$

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 "M.C.Q (1 Marks)" 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

A particle moves so that $S = 6 + 48t - {t^3}$. The direction of motion reverses after moving a distance of
→
The number of real roots of the equation $\mathrm{e}^{4 \mathrm{x}}+2 \mathrm{e}^{3 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}-6=0$ is :
→
Let $A = \{1, 2, 3\}.$ Then the number of relations containing $(1, 2)$ and $(1, 3),$ which are reflexive and symmetric but not transitive is:
→
If the system of linear equations $x - 2y + kz = 1$ ; $2x + y + z = 2$ ;  $3x - y - kz = 3$ Has a solution $(x, y, z) \ne 0$, then $(x, y)$ lies on the straight line whose equation is
→
The area of the region bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates $\text{x}=0,\text{x}=\frac{\pi}{2}$ and the $x-$axis is:
→
Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$, and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals
→
If $\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big],$ then $\lambda+\mu=$
→
The value of $\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$ is:
→
Two functions $f \,\& \,g$ have first $ \&$ second derivatives at $x = 0 \& $ satisfy the relations,$f(0) = \frac{2}{{g(0)}}, f ‘ (0) = 2 g ‘ (0) = 4g (0) , g ‘‘ (0) = 5 f ‘‘ (0) = 6 f(0) = 3$ then :
→
The order of $\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp; }\text{z}\end{bmatrix}$ $\begin{bmatrix}\text{x} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ is:
→