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Exact form, differential equationshey, Ive had a go at this question but im not too sure. :? Solve the following differential equations given in exact formQa) d/dx (yx^2)=x^3Qb)d/dx (ye^x)=e^2x Aa) integrate wrt x. ?d/dx (yx^2)dx=?x^3 dx yx^2=x^4/4+A divide by x^2 to get y---> y=x^2/4 +A Ab) integrate wrt x. ?d/dx(ye^x)dx=?e^2x ye^x=e^2 x^2 /2 divide by e^x to get y---> y=( 1/2(e^2 x^2) )/ e^x Jake |

Re: Exact form, differential equationsLook after the constants, as they help you find general solutions. a) b) I take it that it's definitely e²x and not e^(2x)... |

Re: Exact form, differential equationsthanks for that! for b) it is d/dx (ye^x)= e^(2x)sorry about that. |

Re: Exact form, differential equationsIn that case you get , hence , a family of curves which includes cosh x (for C=½). |

Re: Exact form, differential equationsbrilliant. |

Re: Exact form, differential equationsOk so, By expressing the following differential equations in exact form, find their general solutions. a) e^2x dt/dx + 2e^2x y = x^2 b) x dy/dx + y = x^3 The first one when writing in standard form, dividing by the exponential, do they cancel out leaving e^2x dx as the integrating factor? and for the second one I end up with e^ln(x) .... = x being the integrating factor..? |

Re: Exact form, differential equationsI am assuming a) is: The left side is the differentiation of the product with respect to x: b) That's how I would solve those, however, you are instructed to express the equations in exact form: a) Test for exactness: Since the equation is exact. Thus, we have and integrating with respect to x, we have: Now we take the partial derivative with respect to y: Since we have thus giving: The solution is given implicitly by: thus we have: I'll leave this method for b) to you! :mrgreen: |

Re: Exact form, differential equationsThanks Mark, testing for exactness the process just confirms that the equation is exact, producing the same general solution, when testing for exactness on b) xdy/dx +y=x^3 (y-x^3) dx + x dy =0 Test for exactness d/dy(y-x^3) = 1/2 x^2 d/dx(x) = 1/2 x^2 Is this correct so far |

Re: Exact form, differential equationsNo, you would have: When taking the partial derivative of a function of more than one variable, you treat the other other variables (those you are not differentiating with respect to) as a constant. |

Re: Exact form, differential equationsok then, so I followed your method but I could only get y - 1/4x^4 +C => y=1/4x^4+Cx^-1 , but it does not match the solution as previously stated y=1/4 x^3 +Cx^-1 |

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