Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$ given that $\text{y}=1.$ when $\text{x}=0.$
✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=-4\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=-4\int\text{x dx}$
$\Rightarrow-\frac{1}{\text{y}}=-4\times\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow-\frac{1}{\text{y}}=-2\text{x}^2+\text{C}...(1)$
It is given that at $\text{x}=0,\text{y}=1.$
Substituting the valuse of x and y in (1), we get
$\text{C}=-1$
Therefore, substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=-2\text{x}^2-1$
$\Rightarrow\text{y}=\frac{1}{2\text{x}^2+1}$
Hence, $\text{y}=\frac{1}{2\text{x}^2+1}$ is the required solution.

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

If $\log(\text{x}^2+\text{y}^2)=2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big),$ show that $=\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}-\text{y}}.$
→
Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
→
Find non-zero values of x satisfying the matrix equation:$\text{x}\begin{bmatrix}2\text{x}&2\\3&\text{x}\end{bmatrix}+2\begin{bmatrix}8&5\text{x}\\4&4\text{x}\end{bmatrix}=2\begin{bmatrix}(\text{x}^2+8)&24(10)&6\text{x}\end{bmatrix}$
→
If a, b, c are the langths of sides, BC, CA and AB of a triangle ABC, prove that $\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}=\vec{\text{0}}$ and deduce that $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}.$
→
Find the area enclosed by the curve $y = |x - 1|$ and $y = -|x - 1 | + 1$.
→
Solve the following differential equation:
$\cos^2(\text{x}-2\text{y}) = 1-2\frac{\text{dy}}{\text{dx}}$
→
Solve the following differential equations:$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
→
Show that the four points A, B, C, D with position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},\ \vec{\text{d}}$ respectively such that $3\vec{\text{a}}-2\vec{\text{b}}+5\vec{\text{c}}-6\vec{\text{d}}=0$, are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.
→
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cot^2\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if 0 < x < 1.
→
Solve the following systems of linear equations by cramer's rule:
x - 2y = 4,
-3x + 5y = -7
→