#$&* course mth 279 Query 06 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. * YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the de is not exact confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First rewriting the equation M(t,y)+N(t,y)*y'=0 y'=- ycos(ty)+1/tcos(ty)+2ye^y^2 (tcos(ty)+2ye^y^2)y'=-(ycos(ty)+1) (ycos(ty)+1)+(tcos(ty)+2ye^y^2)y'=0 Now I can see that M(t,y)=ycos(ty)+1 and N(t,y)=tcos(ty)+2ye^y^2 Now I can check and see if this is a exact d.e -tysin(ty)+cos(ty)=-tysin(ty)+cos(ty) H(t,y)=C Now by calculating H by finding the partial anti derivative of M with respect to T H(t,y)=integral sign(ycos(ty)+1)dt =y integral sign cos(ty)dt+integral sign dt =y 1/y sin(ty)+t+p(y) =sin(ty)+t+p(y) H(t,y)=sin(ty)+t+p(y) to find p(y) I will take the partial derivative of H with the respect to y partial/partial y (sin(ty)+t+p(y))=tcos(ty)+2ye^y^2 cos(ty)*t+0+dp(y)/dy=tcos(ty)+2ye^y^2 cos(ty)*t+dp(y)/dy=tcos(ty)+2ye^y^2 dp(y)/dy=2ye^y^2 p(y)=integral 2ye^y^2 p(y)=e^y^2 H(t,y)=sin(ty)+e^y^2+t sin(ty)+e^y^2+t=C NOW I CAN FIND C y applying the intial value sin(pie*0)+e^0^2+pie=C 0+1+pie=C 1+PIE=c sin(ty)+e^y^2+t=pie+1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First I will see if the equation is exact M(t,y)=t^2+y^2 sin(t) partial N/partial t =partial/partial y (t^2+y^2 sin(t)) =0+sin(t) partial/partial y (y^2) =sin(t)*2y now by taking the anti derivative N=integral sign 2ysin(t)dt =2y integral sign sin(t)dt =2y(-cos(t))+p(y) =-2ycos(t)+p(y) M(t,y)=-2ycos(t)+p(y) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:"