17. Consider the nonlinear differential equation dy /dt = y2. (a) Show that y;(t) = 1/(1 t) is a solution. (b) Show that y>() = 2/(1 t) is not a solution. () Why don't these two facts contradict the Linearity Principle? 18. Consider the nonhomogeneous linear equation dy/dt = y + 2. (a) Compute an equilibrium solution for this equation. (b) Verify that y(r) = 2 e is a solution for this equation. (c) Using your results in parts (a) and (b) and the Uniqueness Theorem, explain why the Linearity Principle does not hold for this equation. substitute y, (t) = 2 in DEC . ) 2 (1 - E since ayz - y,hence yz(t)-2 is not a solution to the Int DECI () Linearity principle states that of y (t) is a solution to the homogeneous linal differential equation dy= act)y, then kact) is also a solution to the DE. K is any constant. Theyfore, Linearity property holds true for the linear equations only Here OF dy It = y is a non-linear differential equation. Hence, it contradicts the Linearity principle.17) dy = 42 - (1 ) a The objective is to show that yet ) =_ 1 is a solution of DEC . ) ( 1 - * ) Differentiate y, (t ) = 1 win. F E. ( 1- t ) => - (1- +) (-1) 2 (1-t ) = Theyfore dy, - substitute y, (t ) = 1 inc. 4 . "= ( 18 ) = miz = dy, Hence y, It ) = 1 is a solution of +1 ) [DENY ] 6) The objective is to show that y, It) = 2 is not a solution to 1 -t DE ( I ) : Differentiate yz (t) = 2 wrt t. yz lt ) = d 32 ( 2 ( 1- t ) # ) = - 1( 2) ( 1- t ) " it ( Int ) - - 2 ( 1-+ ) ( - 1) = > 2 ( 1 - + ) 2- 2 use duen Theyore dyz 2 ( 1 - t ) 2ay at = 2- y - ( B ) given differential equation, comparing (A) and ( B) we get, - y = 2 - et : It has been verified dy = - 4+ 2 According to linearity Principle, I y its is a solution of a homogeneay linear equation then any constant multiply with yit ) is a solution of that linear equation. The given equation is nonhomogeneous linear equation, So linearity principle does not hold for this equation18 ) a) the non homogeneous linear equation dy =- 4+ 2 To find the equilibrium solution for this equation, we will squat, day to zero dy alt = - 8+2=0 = ) This is the equilibrium solution for this equation . b ) dy - 4+2 7 dy - dt 2 - Now integrating both sides, 2 - 4 = ) - en ( 2- 4 ) = t+ ( => cis constant. In ( 2- 4 ) = - the 2- 4 = et-c y ( + ) = 2- e (tte ) is a solution of the given equation Now let's verify, y ( t ) = 2 - et is a solution of this equation, 41+ ) = 2 -e - ( A ) differentiating wix. it's dy ! = Of e= dy et