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How do I solve for $x$?

$$\frac{dx}{dt} = 1.4t - 0.5x$$

I know how to do this without the $x$ at the end using integration.

  • 1
    How much of differential equations have you already studied? In particular, have you seen the topics of linear differential equations and of the method of variation of parameters? – 2017-01-03
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    @zipirovich I don't really know the exact name for what I'm studying. But it's the Cambridge A Level. – 2017-01-03
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    @astraTiCon He is saying that DE's of the form $ay'+by+c$ and similar can be easily solved using certain methods. – 2017-01-03

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You can also solve such an ODE pretty elegantly with a substitution. Set $y(t):=0.5x(t)-1.4t$ i.e. $y'=0.5x'-1.4$. And plugging this into the ODE $x'=1.4t-0.5x$ yields

$$2(y'+1.4)=1.4t-(1.4t+y)=-y$$ and therefore $y'=0.5(-y-2.8)$. Now this is pretty standard. Divide by $(-y-2.8)$ and then integrate. You get

$$\int \frac{dy}{-y-2.8}=0.5\int dt$$

I guess you can do it from here. Otherwise just ask. And don't forget to resubstitute to get $x$.

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Hint:

The ODE can be written as: $$\frac{dx}{dt}+0.5x=1.4t$$ Now, an integrating factor is $\mu(t)=\exp\left[\int 0.5~dt\right]=e^{0.5t}$.

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in the first step solve the homogeous equation and for the inhomogeous part make the ansatz $$y_p=At+B$$