Final answer:
To solve the initial value problem dxdy=3x^2+14x with the initial condition y(0)=-6, you can separate variables and integrate. The solution is y(x)=x^2+1-3.
Step-by-step explanation:
To solve the initial value problem dxdy=3x2+14x with the initial condition y(0)=-6, we can separate variables and integrate. We have dxdy=(3x^2+14x)dx. Rearranging the equation, we get dy=dx(3x^2+14x). Now we can separate variables by dividing both sides by (3x^2+14x), giving us dy/(3x^2+14x)=dx. Integrating both sides, we have ∫dy/(3x^2+14x)=∫dx.
On the left side, we can use the substitution u=3x^2+14x, which gives us du=(6x+14)dx. Substituting these values, the left side becomes ∫dy/u=∫(du/(6x+14)).
Now, we can integrate both sides separately. On the left side, we integrate with respect to y, giving us ln|u|+C1, where C1 is the constant of integration. On the right side, we integrate with respect to x, giving us (1/6)ln|6x+14|+C2, where C2 is the constant of integration.
Combining both sides, we have ln|u|=ln|6x+14|/6+C, where C=C2-C1 is the constant of integration. Using the property of logarithms, we can rewrite the equation as |u|=e^((ln|6x+14|/6)+C), which simplifies to |u|=e^(ln|6x+14|/6)*e^C. Since e^(ln|6x+14|/6) is always positive, we can remove the absolute value signs, giving us u=e^(ln|6x+14|/6)*e^C.
Finally, we substitute u=3x^2+14x and simplify the equation to get 3x^2+14x=e^(ln|6x+14|/6)*e^C. Using the properties of exponents, we have 3x^2+14x=(6x+14)*e^C. Now, we can solve for y(x) by substituting y=3x^2+14x in the equation. Rearranging, we have y(x)=(6x+14)*e^C.
Given the initial condition y(0)=-6, we substitute x=0 and y=-6 in the equation to get -6=(6*0+14)*e^C. Simplifying, we have -6=14e^C. Dividing both sides by 14, we get e^C=-6/14. Taking the natural logarithm of both sides, we have C=ln(-6/14).
Finally, substituting C=ln(-6/14) in the equation, we have y(x)=(6x+14)*e^(ln(-6/14)). Simplifying further, we have y(x)=(6x+14)*(-6/14), which simplifies to y(x)=x^2+1-3.