Solve the following differential equation: dy dx -y x = 2x^ 2 - Vidyadip

Question

Solve the following differential equation:

$\frac{dy}{dx} -\frac{y}{x} = 2x^{2}$

✓

Answer

Integrating factor $= e^{-\int\frac{dx}{x}}$

$= e^{-\int\frac{- logx}{}} = \frac{1}{x} $

The solution of different equation is

$ y^\frac{1} {x} = {\int\frac{2x^{2}}{x}} dx + c$

$y^{{}{}} \frac{1}{x} = x^{2} + c$ or

$y = x^{3} + cx$

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 "3 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

If $\text{f}\text{(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},& \text{when}\text{ x}\neq0 \\1,&\text{when} \text{ x}=0\end{cases}$ Find whether f(x) is continuous at x = 0.

If $\text{y}=\frac{\log\text{x}}{\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\log\text{x}-3}{\text{x}^3}.$

Differentiate the following w.r.t. x:
$\tan^{-1}(\sec\text{x}+\tan\text{x}),\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$

If x $\sqrt{1+y}+y \sqrt{1+x}=0$, for $-1 < x < 1$, prove that $\frac{d y}{d x}=-\frac{1}{(1+x)^2}$.

Determine whether $\text{f}(\text{x})=\frac{-\pi}{2}+\sin\text{x}$ is a increasing or decreasing on $\Big(\frac{-\pi}{3},\frac{\pi}{3}\Big).$

Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by $\text{a}\ ^* \ \text{b}=\frac{\text{ab}}{5}$ for all $\text{a, b}\in\text{Q}_0.$ Show that * is commutative as well as associative. Also, find its identity element if it exists.

Differentiate the following w.r.t. x:
$\frac{8^\text{x}}{\text{x}^8}$

If a unit vector $\vec{\text{a}}$ makes an angle $\frac{\pi}{3}$ with $\hat{\text{i}},\frac{\pi}{4}$ with $\hat{\text{j}}$ and an acute angle $\theta$ with $\hat{\text{k}}$, and ,then find the value of $\theta$.

If $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x},$ prove that $\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}.$

Prove the following results:
$\tan^{-1}\frac{2}{3}=\frac{1}{2}\tan^{-1}\frac{12}{5}$