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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation having $y=(\sin^{-1}x)^2+A\cos^{-1}x+B,$where A and B are arbitrary constants,as its general solution.

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Toolbox:
  • A differential equation is a linear differential equation if it is expressible in the form : $ P_0 \Large\frac{d^ny}{dx^n}$$+P_1 \Large\frac{d^{n-1}y}{dx^{n-1}}$$+P_2\Large \frac{d^{n-2}y}{dx^{n-2}}$$+...P_n y=0$ Where $P_0,P_1....$ are constants or functions of the independent of variable x
  • $\large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
  • $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dx}.$$v+\large\frac{dv}{dx}.$$u$
Given $y= (\sin ^{-1}x)^2+A\cos ^{-1}x+B$
Let us first differentiate with respect to x on both sides
$\large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}$$+A\bigg(\large\frac{-1}{\sqrt {1-x^2}}\bigg)+0$
$\large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}-\frac{A}{\sqrt {1-x^2}}$
This can be written as
$\sqrt {1-x^2}.\large\frac{dy}{dx}$$=2 \sin ^{-1}x-A$
Now again differentiating on both sides w.r.t x
Since $\sqrt {1-x^2}.\large\frac{dy}{dx}$ are two functions in the product form, we can apply the product rule $\large\frac{d}{dx}$$(uv)=u.\large\frac{d}{dx}$$(v)+v.\large\frac{d}{dx}$$(u)$
Here let $u=\sqrt {1-x^2},$ hence $ \large\frac{d}{dx}$$(v)=\large\frac{1}{2 \sqrt {1-x^2}}$$(-2x)$
$v=\large\frac{dy}{dx}\qquad \frac{d}{dx}$$(v)=\large\frac{d^2y}{dx^2}$
$\large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
$\sqrt {1-x^2}.\large \frac{d^2y}{dx^2}+\frac{dy}{dx}.\frac{-2x}{2\sqrt {1-x^2}}$$=2 \large\frac{1}{\sqrt {1-x^2}}-0$
$\sqrt {1-x^2}.\large \frac{d^2y}{dx^2}-\frac{x}{\sqrt {1-x^2}}.\frac{dy}{dx}$$= \large\frac{2}{\sqrt {1-x^2}}$
$\Large\frac{(1-x^2)\bigg(\frac{d^2y}{dx^2}\bigg)-x\bigg(\frac{dy}{dx}\bigg)}{\sqrt {1-x^2}}=\frac{2}{\sqrt {1-x^2}}$
=>$(1-x^2)\large\frac{d^2y}{dx^2}$$-x \large\frac{dy}{dx}$$-2=0$
answered May 14, 2013 by meena.p
 

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