Q.3.find the integrating factor and solve the exact solution consider the differential equation (e^(x-y)cosy+cosx+2e^(2x-y)dx + (sinx-e^(x-y)siny+3y^2e^-y)dy Q.5. Prove ?

Answer 1

See below.

#(e^(x-y)cosy+cosx+2e^(2x-y))dx + (sinx-e^(x-y)siny+3y^2e^-y)dy#
We know that #e^y ne 0# so we can simplify the former equation to obtain
#(e^x cosy+e^y cosx+2e^(2x))dx + (e^y sinx-e^x siny+3y^2)dy#

so we have the differential equation

#M(x,y) dx+N(x,y) dy = 0#
Now if #d/(dy)M(x,y) = d/(dx)N(x,y)# then there exists a function #V(x,y)=C# which is the solution in implicit form, to the previous differential equation.

Now

#d/(dy)M(x,y) = -e^x sin y + e^y cos x# #d/(dx)N(x,y) = e^y cos x-e^x sin y#
Then there exist such a #V(x,y)# function

Now

#int_0^x M(x,y) = V(x,y) + phi_1(y)# and #int_0^y N(x,y) = V(x,y)+phi_2(x)#

Integrating we have

#V(x,y) = e^(2x)+e^x cosy+e^y sin x+phi_1(y)# and #V(x,y) = y^3+e^x cos y +e^y sin x+phi_2(x)#

and finally

#phi_1(y) = y^3# and #phi_2(x) = e^(2x)#

hence

#V(x,y) = e^x cosy+e^y sin x+y^3+e^(2x)= C#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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