First verify that yx satisfies the given differential equati

First verify that y(x) satisfies the given differential equation. Then determine the value of the constant C so that y{x) satisfies the given initial condition e^y y\' = 1; y(x) = ln(x + C), y(0) = 0 x dy/dx + 3y = 2x^5·, y(x) = 1/4x^5 + Cx^-3, y(2) = 1 y\' = 3 x^2{y^2 + 1); y(x) = tan(x^3 + C), y(0) = 1.

Solution

a) y(0)=0;

y(x)=ln(x+c)

y\'=1/(x+c)

exp(y). 1/(x+c)=1

exp(ln(x+c))=(x+c)

on substituting we get,

(x+c).1/(x+c)=1

hence the solution satisfies the original differential equation

y(0)=1; gives 0=in(0+c) c=1

b) x.y\'+3.y=2x²

y(x)=1/4.x⁵+C.x^-3

y\'=5/4.x⁴-3.C.x^-4

x.y\'=5/4.x⁵-3.C.x^-3

x.y\'+3.y=5/4.x⁵-3.C.x^-3+3/4.x⁵+3.C.x^-3=8/4.x⁵=2.x⁵=R.H.S

hence the function satisfies the given differential equation.

y(2)=2⁵/4+C/2³=8+C/8

from the given condition in b

8+C/8=1

C=-7/8

c) y\'=3x²(y²+1)

y\'=3x²/(x³+C)

3x²/(x³+C)=3x²/(x³+1)

from the above the value of C=1, if it will satisfy the given y(0)=1 then the given y(x) is the solution of given differential equation.

y(0)=tan(C)=1

C=Pi/4

which is not equal to 1, hence the given solution is not the solution of given differential equation