Finding a general solution to a differential equation, using the integration factor method


Solution 1

We have, $$y'+2xy=e^{-x^2}$$ Compare above ODE with Leibniz equation $\frac{dy}{dx}+P(x)y=Q(x)$, we get, $P(x)=2x $ & $Q(x)=e^{-x^2}$

Now, Integration factor $$I.F.=e^{\int P(x)dx}=e^{\int 2xdx}=e^{x^2}$$ Hence, the general solution is given as $$y(I.F.)=\int Q(x)(I.F.) dx+c$$ $$\implies y(e^{x^2})=\int e^{-x^2}(e^{x^2}) dx+c$$ $$\implies ye^{x^2}=\int dx+c=x+c$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=e^{-x^2}(x+c)}}$$ Where $c$ is the constant of integration

Solution 2

When the DE is in this form: $y'+p(x)y=q(x)$ the integrating factor is $m(x)=e^{\int{p(x)dx}}$.

Solution 3

Integration factor in this case is $e^{x^2}$, multiplied both sides by it:

$$e^{x^2}y'+2xye^{x^2}=1$$ $$(e^{x^2}y)'=1$$






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Elias S
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Elias S

Updated on August 01, 2022


  • Elias S
    Elias S over 1 year

    Use the method of integrating factor to solve the linear ODE

    $$ y' + 2xy = e^{−x^2}.$$

    And verify your answer

    I can solve the ODE as a linear equation (mulitply both sides, subsititute, reverse product rule, integrate etc.) to obtain the answer

    $$ y(x) = c_1 e^{-x^2} e^{-x^2}x $$

    However could someone show me how to do this question using the integrating factor method and (subsequently verifying it using that method?)

    • André Nicolas
      André Nicolas over 8 years
      Multiply by $e^{x^2}$.
  • Elias S
    Elias S over 8 years
    Is there a special integrating factor way of verifying my answer or is it just the same as traditional verification method.
  • Paul Sundheim
    Paul Sundheim over 8 years
    I'm not sure what the difference is between the two methods. In other words, isn't the integrating factor way the traditional way? It isn't difficult to show, given the form above, that the differential form requires the integrating factor $m(x)$.