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Question

Question: How do you solve this differential equation? \[y' + 5y = 0\] ?...

How do you solve this differential equation? y+5y=0y' + 5y = 0 ?

Explanation

Solution

Hint : Here we need to find the general solution. If we observe the given differential equation and on further simplification it is of the form dydx+Py=Q\dfrac{{dy}}{{dx}} + Py = Q . The general solution is given by y.(I.F)=Q.(I.F)dx+cy.\left( {I.F} \right) = \int {Q.(I.F)dx + c} , where I.F is the integrating factor.

Complete step-by-step answer :
Given,
y+5y=0y' + 5y = 0
That is dydx+5y=0\dfrac{{dy}}{{dx}} + 5y = 0
On comparing with dydx+Py=Q\dfrac{{dy}}{{dx}} + Py = Q , we have,
P=5P = 5 and Q=0Q = 0 .
Now we need to find the integrating factor,
I.F=ePdxI.F = {e^{\int {Pdx} }}
On substituting we have,
I.F=e5dxI.F = {e^{\int {5dx} }}
On integrating we have,
I.F=e5xI.F = {e^{5x}}
Thus we have,
I.F=e5x\Rightarrow I.F = {e^{5x}}
We know the general solution is
y.(I.F)=Q.(I.F)dx+cy.\left( {I.F} \right) = \int {Q.(I.F)dx + c}
Then
y.e5x=0.e5xdx+cy.{e^{5x}} = \int {0.{e^{5x}}dx + c}
y.e5x=cy.{e^{5x}} = c
Divide by e5x{e^{5x}} on both sides we have,
y=ce5xy = \dfrac{c}{{{e^{5x}}}}
y=c.e5xy = c.{e^{ - 5x}}
y=c.e5x\Rightarrow y = c.{e^{ - 5x}} . This is the general solution and ‘c’ is the integration constant.
So, the correct answer is “ y=c.e5x \Rightarrow y = c.{e^{ - 5x}} ”.

Note : We know that the integration of xn{x^n} with respect to ‘x’ is xn.dx=xn+1n+1+c\int {{x^n}.dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c , where ‘c’ is the integration constant. In case of indefinite integral we will have integration constant and in definite integral we will have upper limit and lower limit hence we will not have integration content in case of definite integral.