Video Transcript
Solve the differential equation 𝑦 prime plus 𝑥 times 𝑒 to the power of 𝑦 is equal zero.
We’re given a differential equation, and we’re asked to find the solution to this differential equation. The first thing we should do is try and write our differential equation in a form which we know how to solve. So to start, we’ll write 𝑦 prime as d𝑦 by d𝑥. So we’ve rewritten our differential equation as d𝑦 by d𝑥 plus 𝑥 times 𝑒 to the power of 𝑦 is equal to zero. The next thing we could try and do is write our differential equation in the form d𝑦 by d𝑥 is equal to some function of 𝑥. Then we could just integrate both sides of the equation with respect to 𝑥.
However, we have a problem. We have 𝑒 to the power of 𝑦 in our differential equation. And we can’t remove this. So we’ll always going to have this function of 𝑦 in our differential equation. We can see that it’s multiplied by 𝑥. This should remind us of separable differential equations. Remember, we call a differential equation separable if it can be written in the form d𝑦 by d𝑥 is equal to some function 𝑓 of 𝑦 times some function 𝑔 of 𝑥. In our case, we can do this by subtracting 𝑥 times 𝑒 to the power of 𝑦 from both sides of our differential equation. This gives us our function 𝑓 of 𝑦 is 𝑒 to the power of 𝑦 and our function 𝑔 of 𝑥 is negative 𝑥. And the reason we call these separable differential equations is we can try and solve these by separating our variables.
To separate our variables, we’ll start by dividing both sides of the equation through by 𝑒 to the power of 𝑦. Using our laws of exponents, we’ll write this as 𝑒 to the power of negative 𝑦 times d𝑦 by d𝑥 is equal to negative 𝑥. Now, remember, d𝑦 by d𝑥 is not a fraction. However, when we’re solving separable differential equations, we can treat it a little bit like a fraction. This gives us the equivalent statement 𝑒 to the power of negative 𝑦 d𝑦 is equal to negative 𝑥 d𝑥. Then all we need to do is integrate both sides of our equation. We get the integral of 𝑒 to the power of negative 𝑦 with respect to 𝑦 is equal to the integral of negative 𝑥 with respect to 𝑥. We can now evaluate both of these integrals.
Let’s start with the integral of 𝑒 to the power of negative 𝑦 with respect to 𝑦. We recall, for any constant 𝑎 not equal to zero, the integral of 𝑒 to the power of 𝑎𝑦 with respect to 𝑦 is equal to one over 𝑎 𝑒 to the power of 𝑎𝑦 plus the constant of integration 𝐶. We just divide by the coefficient of 𝑦 in our exponent. In our case, the coefficient of 𝑦 is negative one. So we want to divide our integrand by negative one. This gives us negative 𝑒 to the power of negative 𝑦. And we could add a constant of integration here. However, remember, we’re also going to integrate negative 𝑥 with respect to 𝑥. So we’ll just combine these two constants of integration together.
Now, we want to evaluate the integral of negative 𝑥 with respect to 𝑥. Remember, negative 𝑥 is actually negative 𝑥 to the first power. So we can integrate this by using the power rule for integration, which tells us, for any constant 𝑛 not equal to negative one, the integral of 𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐶.
We want to add one to our exponent of 𝑥 and then divide by this new exponent. In our case, the exponent of 𝑥 is one. So we add one to this to get two and then divide by two. This gives us negative 𝑥 squared divided by two plus the constant of integration we’ll call 𝐶 one.
And this is a general solution to our differential equation. However, this is an implicit solution because it’s not of the form 𝑦 is equal to some function of 𝑥. So if possible, we should try to write this in the form 𝑦 is equal to some function of 𝑥. So let’s try and do this. We’ll start by multiplying both sides of our equation through by negative one. And remember, 𝐶 one is a constant. So we can write negative 𝐶 one as a new constant we’ll call 𝐶. So we now have 𝑒 to the power of negative 𝑦 is equal to 𝑥 squared over two plus 𝐶.
Remember, we want to rewrite this equation so 𝑦 is the subject. So we’ll need to take the natural logarithm of both sides of our equation. Doing this, we get the natural logarithm of 𝑒 to the power of negative 𝑦 is equal to the natural logarithm of 𝑥 squared over two plus 𝐶. Now, remember, the natural logarithm function and the exponential function are inverses. So the natural logarithm of 𝑒 to the power of negative 𝑦 would just be equal to negative 𝑦. So we now have negative 𝑦 is equal to the natural logarithm of 𝑥 squared over two plus 𝐶. And the last thing we need to do is multiply both sides of this equation through by negative one.
And this means we’ve written our general solution in the form 𝑦 is equal to some function of 𝑥. Therefore, given the differential equation, 𝑦 prime plus 𝑥 times 𝑒 to the power of 𝑦 is equal to zero, we were able to find the solution 𝑦 is equal to negative one times the natural logarithm of 𝑥 squared over two plus 𝐶.
