Answer:
[tex]y=sinx +Ccos x[/tex]
Step-by-step explanation:
We are given that differential equation
[tex]cosx \frac{dy}{dx}+sinx y=1[/tex]
Divide by cos x on both sides
[tex]\frac{dy}{dx}+\frac{sinx }{cos x}y=\frac{1}{cos x}[/tex]
[tex]\frac{dy}{dx}+tanx y=sec x[/tex]
Using formula:[tex]tanx=\frac{sinx}{cosx },sec x=\frac{1}{cos x}[/tex]
Compare with first order first degree differential equation
[tex]\frac{dy}{dx}+P(x)y=Q(x)[/tex]
We get [tex]P(x)=tanx[/tex]
Q(x)=[tex]sec x[/tex]
I.F=[tex]e^{\int P(x)dx}=e^{\int tanxdx}=e^{lnsecx}=sec x[/tex]
Using formula :[tex]\int tanx dx=ln secx [/tex]
[tex]e^{lna}=a[/tex]
Then solution
[tex]y\times I.F=\int Q(x)\times I.F dx+C[/tex]
Using this
[tex]y\times sec x=\int sec x\times sec xdx+C[/tex]
[tex] y sec x=\int sec^2 xdx+C[/tex]
[tex]y sec x=tanx +C[/tex]
Using formula :[tex]\int sec^2 xdx=tanx [/tex]
[tex]y=\frac{tanx }{sec x}+\frac{C}{sec x}[/tex]
[tex] y=\frac{sinx }{cos x}\times cos x+Ccos x[/tex]
[tex]y=sinx +Ccos x[/tex]
