If y(x) is the solution of the differential equation d y d x + ( 2 x+1 x ) y=e^ -2 x , x0, where y(1)= 1 2 e⁻², then:

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If y(x) is the solution of the differential equation d y d x + ( 2 x+1 x ) y=e^ -2 x , x>0, where y(1)= 1 2 e⁻², then:

Quantrex Academy · Accepted Answer

Correct answer: C. y(x) is decreasing in ( 1 2 , 1 )

Given differential equation is, d y d x + (2+ 1 x ) y=e^ -2 x , x>0 IF =e^ (2+ 1 x ) d x =e^ 2 x+ x =x e^ 2 x Complete solution is given by y(x) x e^ 2 x = x e^ 2 x e^ -2 x d x+c= x d x+cy(x) e^ 2 x x= x² 2 +c Given, y(1)= 1 2 e⁻² 1 2 e⁻² e² 1= 1 2 +c c=0 y(x)= x² 2 e^ -2 x x y(x)= x 2 e^ -2 x Differentiate both sides with respect to xy^ (x)= e^ -2 x 2 (1-2 x) < 0 x ( 1 2 , 1 ) Hence, y(x) is decreasing in ( 1 2 , 1 )

If y(x) is the solution of the differential equation d y d x + ( 2 x+1 x ) y=e^ -2 x , x>0, where y(1)= 1 2 e⁻², then:

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