A curve y = f(x) passes through the point (0, ( 3 4 ) ) . If the slope of the tangent to the curve at any point (x, y) is given by dy dx = e^x e^ 2x + 5e^x + 6 , then the value of f( 2) is equal to

Step-by-step solution

The given differential equation is dy dx = e^x e^ 2x + 5e^x + 6 . Integrating both sides with respect to x : y = e^x (e^x+2)(e^x+3) dx Substitute t = e^x , so dt = e^x dx : y = 1 (t+2)(t+3) dt Using partial fractions: y = ( 1 t+2 - 1 t+3 ) dt y = |t+2| - |t+3| + C y = | e^x+2 e^x+3 | + C The curve passes through (0, ( 3 4 ) ) , so substitute x = 0 and y = ( 3 4 ) : ( 3 4 ) = ( 1+2 1+3 ) + C ( 3 4 ) = ( 3 4 ) + C C = 0 Thus, the function is f(x) = ( e^x+2 e^x+3 ) . To find f( 2) , substitute x = 2 , which gives e^x