Exams Joint Entrance Examination Mathematics Differential Equations Question 84 Question 84 of 490 Medium NUMERICAL
Let y = y ( x ) y=y(x) y = y ( x ) be the solution of the differential equation ( 1 − x 2 ) d y = [ x y + ( x 3 + 2 ) 3 ( 1 − x 2 ) ] d x , − 1 < x < 1 , y ( 0 ) = 0 (1-{x}^{2})dy=[xy+({x}^{3}+2)\sqrt{3(1-{x}^{2})}]dx,-1<x<1,y(0)=0 ( 1 − x 2 ) d y = [ x y + ( x 3 + 2 ) 3 ( 1 − x 2 ) ] d x , − 1 < x < 1 , y ( 0 ) = 0 . If y ( 1 2 ) = m n , m y(\frac{1}{2})=\frac{m}{n},m y ( 2 1 ) = n m , m and n n n are co-prime numbers, then m + n m+n m + n is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 97
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d y d x − x y 1 − x 2 = ( x 3 + 2 ) 3 ( 1 − x 2 ) 1 − x 2 I F = e − ∫ x 1 − x 2 d x = e + 1 2 ln ( 1 − x 2 ) = 1 − x 2 y 1 − x 2 = 3 ∫ ( x 3 + 2 ) d x y 1 − x 2 = 3 ( x 4 4 + 2 x ) + c ⇒ y ( 0 ) = 0 ∴ c = 0 y ( 1 2 ) = 65 32 = m n m + n = 97 \begin{gathered} \frac{dy}{dx}-\frac{xy}{1-{x}^{2}}=\frac{({x}^{3}+2)\sqrt{3(1-{x}^{2})}}{1-{x}^{2}} \\ IF={e}^{-\int \frac{x}{1-{x}^{2}}dx}={e}^{+\frac{1}{2}\ln (1-{x}^{2})}=\sqrt{1-{x}^{2}} \\ y\sqrt{1-{x}^{2}}=\sqrt{3}\int ({x}^{3}+2)dx \\ y\sqrt{1-{x}^{2}}=\sqrt{3}(\frac{{x}^{4}}{4}+2x)+c \\ \Rightarrow y(0)=0\, \therefore c=0 \\ y(\frac{1}{2})=\frac{65}{32}=\frac{m}{n} \\ m+n=97 \end{gathered} d x d y − 1 − x 2 x y = 1 − x 2 ( x 3 + 2 ) 3 ( 1 − x 2 ) I F = e − ∫ 1 − x 2 x d x = e + 2 1 l n ( 1 − x 2 ) = 1 − x 2 y 1 − x 2 = 3 ∫ ( x 3 + 2 ) d x y 1 − x 2 = 3 ( 4 x 4 + 2 x ) + c ⇒ y ( 0 ) = 0 ∴ c = 0 y ( 2 1 ) = 32 65 = n m m + n = 97 Answer the question to unlock