Answer to Question #190578 in Differential Equations for Samiullah jan

Question

a.

cos(x)y'+sin(x)y=2cos^3(x)sin(x)-1 y(π/4)=3√2, 0<=x< π/2

b.

2xy-9x^3+(2y+x^2+1)dy/dx=0 y(0)=-3

c.

y'+4/x y=x^3y^2

Answer 1

a.

$cos(x)y'+sin(x)y=2cos^3(x)(sin(x))^{-1}....................(1)\newline \text{Initial conditions,} y(\frac{π}{4})=3\sqrt{2}, 0<=x< π/2\newline Divide equation (1) by cosx, y'+tan(x)y=2cos^2(x)(sin(x))^{-1}.....................(1)\newline \text{Integrating factor is}\space \newline IF=e^{\int tanx dx}=e^{lnsecx}=secx\newline \text{Therefore, the solution is given by,}\newline y \space secx=\int 2cos^2(x)(sinx)^{-1} secx dx\newline y \space secx=2\int \frac{cosx}{ secx }dx\newline y \space secx=2\int cotx\space dx\newline y \space secx=2ln(sinx)+C\newline y=cosx(2ln(sinx)+C)\newline \text{From initial conditions}\newline y(\frac{π}{4})=3\sqrt{2}=cos(\frac{π}{4})(2ln(sin(\frac{π}{4}))+C)\newline 3\sqrt{2}=(\frac{1}{\sqrt{2}})(2ln(\frac{1}{\sqrt{2}})+C)\newline 3\times 2=ln(\frac{1}{2})+C\newline 6=-ln(2)+C\newline C=6+ln(2)\newline \text{Therefore, solution is}\newline y=cosx(2ln(sinx)+ln(2)+6)$

b.

$2xy-9x^3+(2y+x^2+1)\frac{dy}{dx}=0,\space y(0)=-3\newline (2xy-9x^3)dx-(2y+x^2+1)dy=0 \newline \text{where}\space M=2xy-9x^3, N=-(2y+x^2+1) \newline \text{Now, find partial derivative wrt x and y,} \newline M_{y}=-2x \newline N_{x}=-2x \newline \text{Since,}M_{y}= N_{x}. \newline \text{Therefore, given equation is a exact differential equation.} \newline \text{Integrating first term, taking y as constant } \newline\text{and in second term, integrating those terms}\newline\text{ those are independent of x.} \int(2xy-9x^3)dx-\int(2y+x^2+1)dy=0 \newline x^2y-\frac{9x^4}{4}-y^2-y=c\newline \text{Put initial condition,}\newline c=-24\newline \text{Therefore, required solution is} x^2y-\frac{9x^4}{4}-y^2-y=-24 4x^2(3y^2+10xy-3y+6)=c\newline \newline \text{Thus, general solution of the given differential equation is, }\newline4x^2(3y^2+10xy-3y+6)=c.$

c.

$y'+\frac{4}{x} y=x^3y^2,\space \newline \text{This is a bernoulli equation.} \text{Divide both side by}y^2\newline y^{-2}y'+\frac{4}{x} y^{-1}=x^3,\space \newline Put y^{-1}=z, \implies -y^{-2}\frac{dy}{dx}=\frac{dz}{dx}\newline Therefore, new differential equation is, \newline -z'+\frac{4}{x} z=x^3 \newline z'-\frac{4}{x} z=-x^3 \newline This is a linear differential equation.\newline z\frac{1}{x^4}=\int(-x^3 \times \frac{1}{x^4} )dx \newline z\frac{1}{x^4}=\int\frac{1}{x} dx \newline z\frac{1}{x^4}=lnx+c \newline$