Video: Solve a First Order Differential Equation

Find the solution for the following differential equation for 𝑦(0) = 0: d𝑦 = 𝑒^(π‘₯ + 𝑦) dπ‘₯.

02:50

Video Transcript

Find the solution for the following differential equation for 𝑦 of zero equals zero. d𝑦 equals 𝑒 to the power of π‘₯ add 𝑦 dπ‘₯.

We’re going to need to do some integration here. But remember, in order to do that, we need to have functions of π‘₯ and dπ‘₯ on one side and functions of 𝑦 and d𝑦 on the other side. But how can we separate the variables here when we have a function of π‘₯ and 𝑦? To do this, recall the product rule for exponents. π‘Ž to the power 𝑛 multiplied by π‘Ž to the power π‘š equals π‘Ž to the power of 𝑛 add π‘š. And so, we can rewrite this differential equation as d𝑦 equals 𝑒 to the power π‘₯ multiplied by 𝑒 to the power 𝑦dπ‘₯.

And now, we can separate the variables. To do this, we divide through by 𝑒 to the power 𝑦. So, now, we have functions of 𝑦 and d𝑦 on the left and functions of π‘₯ and dπ‘₯ on the right. So, now, we can integrate both sides. Let’s start by putting this one over 𝑒 to the power 𝑦 in negative exponent form, which is 𝑒 to the power of negative 𝑦. Let’s integrate the right-hand side first. This is the integral of 𝑒 to the power π‘₯ with respect to π‘₯. To do this, recall an important general integral. The integral of 𝑒 to the power π‘˜π‘₯ with respect to π‘₯ equals one over π‘˜π‘’ to the power π‘˜π‘₯ add a constant of integration 𝑐.

So, the integral of 𝑒 to the power π‘₯ is going to be 𝑒 to the power of π‘₯ plus a constant of integration, which I’ll call 𝑐 one. So, now, let’s integrate 𝑒 to the power of negative 𝑦 with respect to 𝑦. We can think of this power negative 𝑦 as negative one 𝑦 if that helps. So, applying the same general integration rule as before, but this time we’ve got 𝑦 and π‘˜ is negative one, we get the 𝑒 to the negative 𝑦 integrates to one over negative one 𝑒 to the power of negative 𝑦 add a constant of integration, which this time we can call 𝑐 two. One over negative one is just negative one. And we can combine our constants of integration 𝑐 one and 𝑐 two with just one constant of integration, which we’ll call 𝑐.

We’re not actually done with this question yet because we were given the condition that 𝑦 of zero equals zero. This means that when π‘₯ equals zero, 𝑦 equals zero. So, we substitute these values into our equation. And we remember that any non-zero number raised to the power of zero is one. So, 𝑒 to the power zero is one. This gives us that negative one equals one add 𝑐. And so, 𝑐 equals negative two. So, we substitute this into our equation. Finally, we can rearrange our equation to get our functions of π‘₯ and 𝑦 on the same side of the equation. This gives us the 𝑒 to the power π‘₯ add 𝑒 to the power of negative 𝑦 equals two.

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