JEE MAIN - Mathematics Hindi (2019 - 9th April Morning Slot - No. 19)
डिफरेंशियल समीकरण का समाधान
$$x{{dy} \over {dx}} + 2y$$ = x2 (x $$ \ne $$ 0) जिसके लिए y(1) = 1, है :
$$x{{dy} \over {dx}} + 2y$$ = x2 (x $$ \ne $$ 0) जिसके लिए y(1) = 1, है :
$$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$$
$$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$$
$$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$
$$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$$
Comments (0)
