Solve the initial value problem : d y d x + y x =2 x + x ^2 x , y (0)=1

The given differential equation is, d y d x +y x=2 x+x^2 x, y(0)=1 It is a linear differential equation. Comparing it with, d y d x +P y=Q P= x, Q=2 x+x^2 x L.F. =e^ p d x =e^ x d x =e^ | x| Solution of the equation is given by, y ( I.F. )= Q ( I.F. ) d x+c y x= (2 x+x^2 x ) d x+c = 2 x x d x+ x^2 x d x+c = 2 x x d x+ [x^2 x x d x- (2 x x d x ) d x ]+c y x= 2 x d x+x^2 x- 2 x x d x+c y x=x^2 x+c (i) Put x =0, y =1 1=0+c c=1 Put c =1 in equation (i), y x=x^2 x+1 y=x^2+1 x y=x^2+ x

Solve the initial value problem : d y d x + y x =2 x + x ^2 x , y…

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Question

Solve the initial value problem : $\frac{d y}{d x}+ y \tan x =2 x + x ^2 \tan x , y (0)=1$

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Answer

The given differential equation is,
$\frac{d y}{d x}+y \tan x=2 x+x^2 \tan x, y(0)=1$
It is a linear differential equation. Comparing it with,
$\frac{d y}{d x}+P y=Q$
$P=\tan x, Q=2 x+x^2 \tan x$
$\text { L.F. }=e^{\int p d x}$
$=e^{\int \tan x d x}$
$=e^{\log |\sec x|}$
Solution of the equation is given by,
$y \times(\text { I.F. })=\int Q \times(\text { I.F. }) d x+c$
$y \sec x=\int\left(2 x+x^2 \tan x\right) \sec d\ x+c$
$=\int 2 x \sec x \ d x+\int x^2 \sec \tan x\ d x+c$
$=\int 2 x \times \sec x\ d x+\left[x^2 \times \int \sec x \tan x\ d x-\int\left(2 x \int \sec \tan x\ d x\right) d x\right]+c$
$y \sec x=\int 2 x \sec d x+x^2 \sec x-\int 2 x \sec x\ d x+c$
$y \sec x=x^2 \sec x+c \ldots \text { (i) }$
Put $x =0, y =1$
$1=0+c$
$c=1$
Put $c =1$ in equation $(i),$
$y \sec x=x^2 \sec x+1$
$y=x^2+\frac{1}{\sec x}$
$y=x^2+\cos x$