What is the solution to the Differential Equation #e^(x+y)(dy/dx) = x# with #y(0)=1?
# y = ln (xe^x  e^x + e+1) #
We have:
We can rearrange the DIfferential Equation [A] as follows:
This is a First Order Separable Differential equation ad we now "seperate the variables" to get:
The LHS integral is a standard result, and for the RHS integral we would need to apply Integration By Parts:
Then plugging into the IBP formula:
gives us
Using this result, we can now integrate [B] to get the General Solution:
Hence, the Particular Solution is:
And we can gain an explicit solution for [A] if we take Natural Logarithms:
So that finally:
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To solve the differential equation ( e^{x+y} \frac{dy}{dx} = x ) with the initial condition ( y(0) = 1 ), we can use the method of separation of variables.

Rearrange the equation to isolate ( \frac{dy}{dx} ): ( \frac{dy}{dx} = \frac{x}{e^{x+y}} )

Separate variables: ( e^{y} dy = x e^{x} dx )

Integrate both sides: ( \int e^{y} dy = \int x e^{x} dx ) Integrating ( e^{y} ) with respect to ( y ) gives ( e^{y} ). Integrating ( x e^{x} ) with respect to ( x ) requires integration by parts: Let ( u = x ) and ( dv = e^{x} dx ), then ( du = dx ) and ( v = e^{x} ). Using integration by parts, we get: ( \int x e^{x} dx = xe^{x}  \int e^{x} dx = xe^{x} + e^{x} + C_1 ), where ( C_1 ) is the constant of integration.

Substitute back and simplify: ( e^{y} = xe^{x} + e^{x} + C_1 ) ( y = \ln(xe^{x} + e^{x} + C_1) )

Apply the initial condition ( y(0) = 1 ) to find ( C_1 ): ( 1 = \ln(0e^{0} + e^{0} + C_1) ) ( 1 = \ln(1 + C_1) ) ( e = 1 + C_1 ) ( C_1 = e  1 )

Substitute ( C_1 ) back into the solution: ( y = \ln(xe^{x} + e^{x} + e  1) )
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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