# Answer to Question #6732 in Differential Equations for josh

Question #6732

dy/dx - ay = Q, where a is a constant, giving your snswer in terms of of a, when

a) Q = ke^lamda(x)

b) Q = ke^ax

c) Q = (kx^n)(e^ax)

12) use the substitution z = y^-1 to transform the differntial equation x dy/dx +y = (y^2)ln(x), into a linear equation. hense obtain the general solution of the original equation.

13) use the substitution z = y^2 to transform the differential equation 2cosx dy/dx - ysinx +y^-1 = 0, into a linear equation. hense obtain the general solution of the original equation.

14) us the substitution u = x + y to transform the differential equation dy/dx = (x +y +1)(x + y - 1) into a differential equation in u and x. by first solving this new equation. hense obtain the general solution of the original equation

15) use the substitution u = y - x - 2 to transform the differential equation dy/dx = (y - x - 2)^2 into a differntial equation in u and x. By first solving this new equation, find the general solution of the original equation, giving y in therms of x.

a) Q = ke^lamda(x)

b) Q = ke^ax

c) Q = (kx^n)(e^ax)

12) use the substitution z = y^-1 to transform the differntial equation x dy/dx +y = (y^2)ln(x), into a linear equation. hense obtain the general solution of the original equation.

13) use the substitution z = y^2 to transform the differential equation 2cosx dy/dx - ysinx +y^-1 = 0, into a linear equation. hense obtain the general solution of the original equation.

14) us the substitution u = x + y to transform the differential equation dy/dx = (x +y +1)(x + y - 1) into a differential equation in u and x. by first solving this new equation. hense obtain the general solution of the original equation

15) use the substitution u = y - x - 2 to transform the differential equation dy/dx = (y - x - 2)^2 into a differntial equation in u and x. By first solving this new equation, find the general solution of the original equation, giving y in therms of x.

Expert's answer

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment