Let y = y(x) be the solution of the differential
equation (x $-$ x3)dy = (y + yx2 $-$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
Solution
$(x - {x^3})dy = (y + y{x^2} - 3{x^4})dx$<br><br>$\Rightarrow xdy - ydx = (y{x^2} - 3{x^4})dx + {x^3}dy$<br><br>$\Rightarrow {{xdy - ydx} \over {{x^2}}} = (ydx + xdy) - 3{x^2}dx$<br><br>$\Rightarrow d\left( {{y \over x}} \right) = d(xy) - d({x^3})$<br><br>Integrate<br><br>$\Rightarrow {y \over x} = xy - {x^3} + c$<br><br>given f(3) = 3<br><br>$\Rightarrow {3 \over 3} = 3 \times 3 - {3^3} + c$<br><br>$\Rightarrow c = 19$<br><br>$\therefore$ ${y \over x} = xy - {x^3} + 19$<br><br>at $x = 4,{y \over 4} = 4y - 64 + 19$<br><br>$15y = 4 \times 45$<br><br>$\Rightarrow y = 12$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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