Solve the following differential equation 1-x^4 dy=x dx - Vidyadip

Question

Solve the following differential equation
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$

✓

Answer

We have,
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$
$\Rightarrow\text{dy}=\frac{\text{x}}{\sqrt{1-\text{x}^4}}\ \text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int\frac{\text{x}}{\sqrt{1-\text{x}^4}}\text{ dx}$
$\Rightarrow\text{y}=\int\frac{\text{x}}{\sqrt{1-\text{x}^4}}\ \text{dx}$
Putting $x^2= t$
$\Rightarrow2\text{x dx}=\text{dt}$
$\therefore\text{y}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$=\frac{\sin^{-1}}{2}+\text{C}$
$=\frac{1}{2}\sin^{-1}(\text{x}^2)+\text{C}$
hence, $\text{y}=\frac{1}{2}\sin^{-1}(\text{x}^2)+\text{C}$ is the solution to the given differential equation.

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

Differentiate the following functions with respect to x:
$\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$

If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{If x}=\sqrt{\text{a}^{\sin^{-1}}\text{t}},\text{y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}},\text{ Show that}\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$

If $x = \sin \text{t} $ and $\text{y} = \sin \text{pt,}$ prove that $(1 - x^{2}) \frac{\text{d}^{2} \text{y}}{\text{dx}^{2}} - x \frac{\text{dy}}{\text{dx}} + \text{P}^{2}\text{y} = 0.$

Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{\text{x}}\sin{\text{x}}\text{ on }[0,\pi]$

Integrate the function in Exercise:
$\text{e}^\text{x}\Bigg(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Bigg)$

Show that the points $\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$ are equidistant from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$

If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find0 $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}\Big(\cos\text{t}+\log\tan\frac{\text{t}}{2}\Big)\text{y}=\text{a}\sin\text{t}$

Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$

Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1, x = 4$ and the x-axis.

Solve $\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2.$