. The general solution of the differential equation

y' + yϕ' (x) - ϕ (x) . ϕ' (x) = 0 where ϕ(x) is a known function is

where c is an arbitrary constant.

A: y = ce-ϕ(x) + ϕ (x) - 1

B: y = ce+ϕ(x) + ϕ (x) - 1

C: y = ce-ϕ(x) - ϕ (x) + 1

D: y = ce-ϕ(x) + ϕ (x) + 1

Best Answer


We know that if the equation is dy/dx+P(x)y=Q(x) then the integrating factor =e∫P(x)dx

Given the differential equation is 

y'+y ϕ'(x)- ϕ(x). ϕ'(x)=0

⇒y'+ ϕ'(x)y= ϕ(x). ϕ'(x)

Then integrating factor =e∫ϕ'(x)dx=eϕ(x)

Multiplying both sides by integrating factor we get, 

ϕ(x)y'+e ϕ(x)' ϕ(x)y=e ϕ(x)(x) ϕ'(x)

⇒d(y.e ϕ(x))=e ϕ(x)(x) ϕ'(x)

Integrating both sides,

∫d(y.e ϕ(x))=∫e ϕ(x) ϕ(x) ϕ'(x)dx

⇒ye ϕ(x)=ϕ(x)∫eϕ(x)∫ϕ'(x)dx-∫(d( ϕ(x))/dx ×∫eϕ(x) ϕ'(x)dx)dx

⇒ye ϕ(x)= ϕ(x)eϕ(x)-∫ϕ'(x)e ϕ(x)dx

⇒ye ϕ(x)= ϕ(x)e ϕ(x)-e ϕ(x)+c, where c is an arbitrary constant 

⇒y= ϕ(x)-1+ce- ϕ(x), multiplying both sides by e- ϕ(x)

Final Answer:

Hence the correct option is (A). i.e y=ce -ϕ(x)+ ϕ(x)-1

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