Find General Solution of dy/dx+ytanx = secx

The general solution of the differential equation dy/dx+ytanx = secx is equal to ysecx =tanx +C where C is a constant. This equation is an example of a first order linear ordinary differential equation.

Solve dy/dx+ytanx = secx

Question: Solve the differential equation $\dfrac{dy}{dx}$+ytanx = secx.

Solution:

The given equation has the form

$\dfrac{dy}{dx}+P(x)=Q(x)$,

a 1st order linear ordinary differential equation (ode). So the given equation is a first order linear ode with

P(x) = tanx, Q(x) =secx.

Its integrating factor I(x) = e^∫P(x)dx = e^{∫tanx dx} = e^{log secx} = secx.

So the solution will be

y I(x) = ∫Q(x) I(x) dx +C

⇒ y secx = ∫secx secx dx +C

⇒ y secx = ∫sec²x dx +C

⇒ y secx = tanx dx +C

So the general solution of the ode dy/dx+ytanx = secx is equal to y secx = tanx dx +C where C is an arbitrary integral constant.

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Solve (x+y+1)dy/dx=1

Solve dy/dx =sin(x+y)

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