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x(x^2-1)(dy)/(dx)=1; y=0w h e nx=2...

`x(x^2-1)(dy)/(dx)=1; y=0w h e nx=2`

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Find the particular solution of the differential equation x(x^2-a)(dy)/(dx)=1;\ \ \ \ \ y=0\ \ "w h e n"\ x=2

Find a particular solution of the differential equation: (x+y)dy+(x-y)dx=0; y=1\ w h e n\ x=1

If e^(x) + e^(y) = e^(x + y) , then prove that (dy)/(dx) = (e^(x)(e^(y) - 1))/(e^(y)(e^(x) - 1)) or (dy)/(dx) + e^(y - x) = 0 .

For each of the following initial value problems verify that the accompanying functions is a solution. (i) x(dy)/(dx)=1, y(1)=0 => y=logx (ii) (dy)/(dx)=y , y(0)=1 => y=e^x (iii) (d^2y)/(dx^2)+y=0, y(0)=0, y^(prime)(0)=1 => y=sinx (iv) (d^2y)/(dx^2)-(dy)/(dx)=0, y(0)=2, y^(prime)(0)=1 => y=e^x+1 (v) (dy)/(dx)+y=2, y(0)=3 => y=e^(-x)+2

(x+1)(dy)/(dx) -1 = 2e^(-y) , y=0, " when " x=1

(1+x^(2))(dy)/(dx)+y=e^(tan^(-1)x)

(1+x^(2))(dy)/(dx)+y=e^(tan^(-1)x)

(1+x^(2))(dy)/(dx)+y=e^(tan^(-1)x)

(1+x^(2))(dy)/(dx)+y=e^(tan^(-1)x)