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I can't solve these differential, someone can help me with a step by step solution? thanks $$y'+ty=t^3$$ $$y'=3t^2y+4t^2$$

I tried the first integrating by $$e^{\int tdt}$$ using $$p(t)=t$$ and $$q(t)=t^3$$

so I have $$ye^{\int\ tdt} = {\int}t^3e^{\int\ tdt}$$

but then I don't know how to proceed

Paolo
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3 Answers3

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Use the integrating factor method. See the answer to your earlier question Differential Equation $ty'= 3t^2-y$ solution incorrect $$\frac{dy}{dx}+P(x)y=Q(x) \Rightarrow I=e^{\int P(x)dx}$$ Then $$yI=\int IQdx +c$$

Community
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David Quinn
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$$ \frac{dy}{dt} + ty = t^3$$

Using Integrating factor method here:

Integrating factor $e^{\int t dt}= e^{\frac{t^2}{2}}$

Multiply through by integrating factor $$e^{\frac{t^2}{2}}\frac{dy}{dt} + tye^{\frac{t^2}{2}} = t^3e^{\frac{t^2}{2}} $$

By reverse chain rule:

$$ \frac{d}{dt} ye^{\frac{t^2}{2}} = t^3e^{\frac{t^2}{2}}$$

Integrating both sides wrt t:

$$ ye^{\frac{t^2}{2}} = \int t^3e^{\frac{t^2}{2}}\ dt $$

To integrate RHS:

Let $u = \frac{t^2}{2}$,

$$\frac{du}{dt} = 2t $$

$$\int t^3e^{\frac{t^2}{2}} = \frac{1}{2} \int ue^{\frac{u}{2}} \ du $$

You can now integrate by parts to find this integral and solve for y:

$$ \int ue^{\frac{u}{2}} \ du$$

Let $v = u$, $\frac{dp}{du} = e^{\frac{u}{2}} $

$\frac{dv}{du} = 1 $, $p = 2e^{\frac{u}{2}} $

$\int ue^{\frac{u}{2}} \ du = 2e^{\frac{u}{2}}(u-2) + c $

Can you finish it off now?

Please use the same technique to solve the second equation. To read more, see the following link

John_dydx
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  • thank you, wasn't simple for me to continue the equation after my part – Paolo Aug 22 '15 at 13:20
  • You're very welcome, if you would like me to clarify anything, please let me know. I'm very happy to help if I see you've made some effort which I think you have. – John_dydx Aug 22 '15 at 13:22
  • can you clarify the integration of $$1/(3y+4)$$

    isn't the result $$log(3y+4) +c $$ ?

     – Paolo Aug 22 '15 at 13:42
  • The integral of $\frac{1}{f(x)}= \frac{1}{f'(x)}\ln|f(x)|$, where f(x)=ax+b. Make sense?See this link for more-http://mymathforum.com/calculus/8203-what-technique-use-get-antiderivative-1-2x-1-a.html. You need to understand integration and differentiation really well to solve differential equations. – John_dydx Aug 22 '15 at 13:47
  • ok , I've a formulary for infinite but this integral is not included – Paolo Aug 22 '15 at 13:51
  • It doesn't apply for every function of x-for example $\int \frac{1}{x^2} \ne \ln|x^2| + c$, However, $\int \frac{1}{ax+b} = \frac{1}{a}\ln|ax+b| + c$. See this link for more-http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Integrals_Reduced.pdf – John_dydx Aug 22 '15 at 13:56
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Notice, we have $$\frac{dy}{dt}+ty=t^3$$ $$I.F.=e^{\int tdt}=e^{t^2/2}$$ Hence, the solution is given as $$y(I.F.)=\int t^3(I.F.)dt+c$$ $$ye^{t^2/2}=\int t^3 e^{t^2/2}dt+c$$ Let $\frac{t^2}{2}=u\implies tdt=du$ $$ye^{t^2/2}=\int 2u e^{u}du+c$$ $$ye^{t^2/2}=2u\int e^{u}du-2\int e^udu+c$$ $$ye^{t^2/2}= 2e^{u}(u-1)+c$$

$$ye^{t^2/2}=2e^{t^2/2}\left(\frac{t^2}{2}-1\right)+c$$

Similarly, $$\frac{dy}{dt}=3t^2y+4t^2$$ $$\frac{dy}{dt}=t^2(3y+4)$$ $$\frac{dy}{3y+4}=t^2dt$$ $$\int \frac{dy}{3y+4}=\int t^2dt$$ $$\frac{1}{3}\ln|3y+4|=\frac{t^3}{3}+c$$

Harish Chandra Rajpoot
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