a) (3x^2+3xy^2)dx+(3x^2-3y^2+2y)dy=0 Integrated both with respect to x and y x^3

Question

a) (3x^2+3xy^2)dx+(3x^2-3y^2+2y)dy=0 Integrated both with respect to x and y x^3+(3/2)x^2y^2+c+(3/2)x^2y^2-y^3+y^2=0 But the answer is x^3+(3/2)x^2y^2-y^3+y^2=0 b) y^3sinx dx - 3y^2cos^2(x)=0 Integrating both i got -y^3cosx-y^3cos^2(x) The answer is y^3(1+cos2x)

Explanation / Answer

(a)to get the answer, i guess there is a slight mistake in the question: (3x^2+3xy^2)dx+(3x^2y-3y^2+2y)dy=0 there should be 3x^2 y dy instead of 3x^2 in the second parantheses. now, the sol becomes: ==> 3x^2dx + 3(xy^2 dx + x^2y dy) - 3y^2dy + 2ydy = 0 ==> d(x^3) + 1.5(2xy^2 dx + 2x^2y dy) - d(y^3) + d(y^2) = 0 ==> d(x^3) + 1.5 d(x^2y^2) - d(y^3) + d(y^2) = 0 ==> d(x^3 + 1.5 x^2y^2 - y^3 + y^2) = 0 integrating, we get: x^3 + 1.5 x^2y^2 - y^3 + y^2 = constant (b) there is a mistake in this part as well; y^3sin2x dx - 3y^2(cosx)^2 dy=0, it should be sin 2x instead of sin x in the first part ==> 2y^3 sinx cosx dx - 3y^2(cosx)^2 dy=0 ==> y^3 d(-(cosx)^2) - (cosx)^2 d(y^3) = 0 ==> y^3 d(-(cosx)^2) + (-(cosx)^2) d(y^3) = 0 ==> d(-y^3(cosx)^2) = 0 integrating, we get: y^3(cosx)^2 = constant ==> y^3 x 2(cosx)^2 = constant ==> y^3(1 + cos 2x) = constant

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