Exams Joint Entrance Examination Mathematics Differential Equations Question 71 Question 71 of 490 Medium NUMERICAL
If y = y ( x ) y=y(x) y = y ( x ) is the solution of the differential equation,
4 − x 2 d y d x = ( ( sin − 1 ( x 2 ) ) 2 − y ) sin − 1 ( x 2 ) , − 2 ≤ x ≤ 2 , y ( 2 ) = π 2 − 8 4 \sqrt{4-{x}^{2}}\frac{dy}{dx}=({({\sin }^{-1}(\frac{x}{2}))}^{2}-y){\sin }^{-1}(\frac{x}{2}),-2\le x\le 2,y(2)=\frac{{\pi }^{2}-8}{4} 4 − x 2 d x d y = ( ( sin − 1 ( 2 x )) 2 − y ) sin − 1 ( 2 x ) , − 2 ≤ x ≤ 2 , y ( 2 ) = 4 π 2 − 8 , then y 2 ( 0 ) {y}^{2}(0) y 2 ( 0 ) is equal to \_\_\_\_\_\_\_\_\_\_\_.
Answer: 4
Solutions · 1 Post your solution Official solution
d y d x + ( sin − 1 x 2 ) 4 − x 2 y = ( sin − 3 x 2 ) 3 4 − x 2 y e ( sin − 1 x 2 ) 2 2 = ∫ ( sin − 3 x 2 ) 3 4 − x 2 e ( sin − 1 x 2 ) 2 2 d x y = ( sin − 1 x 2 ) 2 − 2 + c ⋅ e − ( sin − 1 x 2 ) 2 2 y ( 2 ) = π 2 4 − 2 ⇒ c = 0 y ( 0 ) = − 2 \begin{gathered} \frac{dy}{dx}+\frac{({\sin }^{-1}\frac{x}{2})}{\sqrt{4-{x}^{2}}}y=\frac{{({\sin }^{-3}\frac{x}{2})}^{3}}{\sqrt{4-{x}^{2}}} \\ y{e}^{\frac{{({\sin }^{-1}\frac{x}{2})}^{2}}{2}}=\int \frac{{({\sin }^{-3}\frac{x}{2})}^{3}}{4-{x}^{2}}{e}^{\frac{{({\sin }^{-1}\frac{x}{2})}^{2}}{2}}dx \\ y={({\sin }^{-1}\frac{x}{2})}^{2}-2+c\cdot {e}^{\frac{-{({\sin }^{-1}\frac{x}{2})}^{2}}{2}} \\ y(2)=\frac{{\pi }^{2}}{4}-2\Rightarrow c=0 \\ y(0)=-2 \end{gathered} d x d y + 4 − x 2 ( sin − 1 2 x ) y = 4 − x 2 ( sin − 3 2 x ) 3 y e 2 ( s i n − 1 2 x ) 2 = ∫ 4 − x 2 ( sin − 3 2 x ) 3 e 2 ( s i n − 1 2 x ) 2 d x y = ( sin − 1 2 x ) 2 − 2 + c ⋅ e 2 − ( s i n − 1 2 x ) 2 y ( 2 ) = 4 π 2 − 2 ⇒ c = 0 y ( 0 ) = − 2 Answer the question to unlock