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If dy/dx = x^2 + 2, then find the value of y in terms of x. It is given that initially, x = 0 and y = 3.
Hi dy/dx = x^2 + 2 integrating both sides w.r.t x, we get each term as follows: Integration of dy/dx = y + a x^2 = (x^3)/3 + b 2 = 2x + c (a,b and c are constants of integration) Now, combining the terms and taking a,b and c on one side of equation and replacing the resulting constant as Z we get: y= (x^3)/3 + 2x + Z But we are given that y=3 when x=0 putting these values in above equation: 3=0 + 0 + Z Z = 3 Hence, we get our answer as : y= (x^3)/3 + 2x + 3
Hi
dy/dx = x^2 + 2 integrating both sides w.r.t x, we get each term as follows:
Integration of dy/dx = y + a x^2 = (x^3)/3 + b 2 = 2x + c (a,b and c are constants of integration)
Now, combining the terms and taking a,b and c on one side of equation and replacing the resulting constant as Z we get:
y= (x^3)/3 + 2x + Z
But we are given that y=3 when x=0 putting these values in above equation:
3=0 + 0 + Z Z = 3
Hence, we get our answer as : y= (x^3)/3 + 2x + 3
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