How do you do (a) and (b)?Bernoulli’s equation is an equation of the form y ′ = a(t)y + f(t)y n , where n can't be 0 or 1.(a) Using the substitution z = y 1−n , show that we can transform Bernoulli’s equation into the linear equation z ′ = (1 − n)a(t)z + (1 − n)f(t).(b) Solve the equation xy′ + y = x^4 y^3

Answer:See solution belowStep-by-step explanation:(a) If n=0 or 1, the equation(1)  [tex] y' = a(t)y + f(t)y^n [/tex]would be a simple linear differential equation. So, we can assume that n is different  to 0 or 1.Let's use the following substitution:(2) [tex]z=y^{n-1}[/tex]Taking the derivative implicitly and using the chain rule:(3) [tex]z'=(1-n)y^{-n}y'[/tex]Multiplying equation (1) on both sides by[tex](1-n)y^{-n}[/tex]we obtain the equation[tex](1-n)y^{-n}y' = (1-n)y^{-n}a(t)y+(1-n)y^{-n}f(t)y^n[/tex]reordering:[tex](1-n)y^{-n}y' = (1-n)y^{-n}ya(t)+(1-n)y^{-n}y^nf(t)[/tex][tex](1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)y^{0}f(t)[/tex][tex](1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)f(t)[/tex]Now, using (2) and (3) we get:[tex]z'= (1-n)za(t)+(1-n)f(t)[/tex]which is an ordinary linear differential equation with unknown function z(t).(b)The equation we want to solve is(4)   [tex]xy'+ y = x^4 y^3[/tex]  Here, our independent variable is x (instead of t)Assuming x different to 0, we divide both sides by x to obtain:[tex]y'+\frac{1}{x}y = x^3 y^3[/tex][tex]y' = -\frac{1}{x}y+x^3 y^3[/tex]Which is an equation of the form (1) with[tex]a(x)=-\frac{1}{x}[/tex][tex]f(x)=x^3[/tex][tex]n=3[/tex]So, if we substitute[tex]z=y^{-2}[/tex]we transform equation (4) in the lineal equation(5) [tex]z'=\frac{2}{x}z-2x^3[/tex]and this is an ordinary lineal differential equation of first order whoseintegrating factor is[tex]e^{\int (-\frac{2}{x})dx}[/tex]but[tex]e^{\int (-\frac{2}{x})dx}=e^{-2\int \frac{dx}{x}}=e^{-2ln(x)}=e^{ln(x^{-2})}=x^{-2}=\frac{1}{x^2}[/tex]Similarly,[tex]e^{\int (\frac{2}{x})dx}=x^2[/tex]and the general solution of (5) is then[tex]z(x)=x^2\int (\frac{-2x^3}{x^2})dx+Cx^2=-2x^2\int xdx+Cx^2=\\\ -2x^2\frac{x^2}{2}+Cx^2=-x^4+Cx^2[/tex]where C is any real constantReversing the substitution  [tex]z=y^{-2}[/tex]we obtain the general solution of (4)[tex]y=\sqrt{\frac{1}{z}}=\sqrt{\frac{1}{-x^4+Cx^2}}[/tex]Attached there is a sketch of several particular solutions corresponding to C=1,4,6It is worth noticing that the solutions are not defined on x=0 and for C<0