By introducing a new variable t- putting x-cos t- the expression 1-x2 d2 y-d x2-x d y-d x-y is transformed into -A- d2 y-d t2-yB- d2 y-d t2-t d y-d t-yC- d2 y-d t2-yD- d2 y-d t2

The correct option is A t=cos^ 1x⇒dt/dx= 1/√(1 x^2) dy/dx=dy/dx.dt/dx= 1/√(1 x^2).dy/dt⇒ dy/dt ⇒ (1 x^2)d^2y/dx^2 xdy/dx=d^2y/dt^2 ...

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Standard XII

Mathematics

Integration by Substitution

By introducin...

Question

By introducing a new variable t, putting x = cos t, the expression $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + y$ is transformed into :

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Solution

The correct option is A
$t = c o s^{- 1} x \Rightarrow \frac{d t}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \frac{d y}{d x} = \frac{d y}{d x} . \frac{d t}{d x} = - \frac{1}{\sqrt{1 - x^{2}}} . \frac{d y}{d t} \Rightarrow - \frac{d y}{d t} \Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = \frac{d^{2} y}{d t^{2}}$

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By introducing a new variable t, putting x = cos t, the expression (1−x2)d2ydx2−xdydx+y is transformed into :

The correct option is A t=cos−1x⇒dtdx=−1√1−x2dydx=dydx.dtdx=−1√1−x2.dydt⇒−dydt⇒(1−x2)d2ydx2−xdydx=d2ydt2

By introducing a new variable t, putting x = cos t, the expression $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + y$ is transformed into :